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3.5: Implicit Differentiation - Mathematics


In the previous sections we learned to find the derivative, ( frac{dy}{dx}), or (y^prime ), when (y) is given explicitly as a function of (x). Knowing (x), we can directly find (y).)

Sometimes the relationship between (y) and (x) is not explicit; rather, it is implicit. For instance, we might know that (x^2-y=4). This equality defines a relationship between (x) and (y); if we know (x), we could figure out (y). Can we still find (y^prime )? In this case, sure; we solve for (y) to get (y=x^2-4) (hence we now know (y) explicitly) and then differentiate to get (y^prime =2x).

Sometimes the implicit relationship between (x) and (y) is complicated. Suppose we are given (sin(y)+y^3=6-x^3). A graph of this implicit function is given in Figure 2.19. In this case there is absolutely no way to solve for (y) in terms of elementary functions. The surprising thing is, however, that we can still find (y^prime ) via a process known as implicit differentiation.

Implicit differentiation is a technique based on the Chain Rule that is used to find a derivative when the relationship between the variables is given implicitly rather than explicitly (solved for one variable in terms of the other).

We begin by reviewing the Chain Rule. Let (f) and (g) be functions of (x). Then [frac{d}{dx}Big(f(g(x))Big) = f^prime(g(x))cdot g'(x).] Suppose now that (y=g(x)). We can rewrite the above as [frac{d}{dx}Big(f(y))Big) = f^prime(y))cdot y^prime , quad ext{or}quad frac{d}{dx}Big(f(y))Big)= f^prime(y)cdot frac{dy}{dx}.label{2.1} ag{2.1}] These equations look strange; the key concept to learn here is that we can find (y^prime ) even if we don't exactly know how (y) and (x) relate.

We demonstrate this process in the following example.

Example 67: Using Implicit Differentiation

Find (y^prime ) given that (sin(y) + y^3=6-x^3).

Solution

We start by taking the derivative of both sides (thus maintaining the equality.) We have :

[ frac{d}{dx}Big(sin(y) + y^3Big)=frac{d}{dx}Big(6-x^3Big).]

The right hand side is easy; it returns (-3x^2).

The left hand side requires more consideration. We take the derivative term--by--term. Using the technique derived from Equation 2.1 above, we can see that [frac{d}{dx}Big(sin yBig) = cos y cdot y^prime .]

We apply the same process to the (y^3) term.

[frac{d}{dx}Big(y^3Big) = frac{d}{dx}Big((y)^3Big) = 3(y)^2cdot y^prime .]

Putting this together with the right hand side, we have

[cos(y)y^prime +3y^2y^prime = -3x^2.]

Now solve for (y^prime ).

[egin{align*} cos(y)y^prime +3y^2y^prime &= -3x^2. ig(cos y+3y^2ig)y^prime &= -3x^2 y^prime &= frac{-3x^2}{cos y+3y^2} end{align*}]

This equation for (y^prime ) probably seems unusual for it contains both (x) and (y) terms. How is it to be used? We'll address that next.

Implicit functions are generally harder to deal with than explicit functions. With an explicit function, given an (x) value, we have an explicit formula for computing the corresponding (y) value. With an implicit function, one often has to find (x) and (y) values at the same time that satisfy the equation. It is much easier to demonstrate that a given point satisfies the equation than to actually find such a point.

For instance, we can affirm easily that the point ((sqrt[3]{6},0)) lies on the graph of the implicit function (sin y + y^3=6-x^3). Plugging in (0) for (y), we see the left hand side is (0). Setting (x=sqrt[3]6), we see the right hand side is also (0); the equation is satisfied. The following example finds the equation of the tangent line to this function at this point.

Example 68: Using Implicit Differentiation to find a tangent line

Find the equation of the line tangent to the curve of the implicitly defined function (sin y + y^3=6-x^3) at the point ((sqrt[3]6,0)).

Solution

In Example 67 we found that [y^prime = frac{-3x^2}{cos y +3y^2}.]We find the slope of the tangent line at the point ((sqrt[3]6,0)) by substituting (sqrt[3]6) for (x) and (0) for (y). Thus at the point ((sqrt[3]6,0)), we have the slope as [y^prime = frac{-3(sqrt[3]{6})^2}{cos 0 + 3cdot0^2} = frac{-3sqrt[3]{36}}{1} approx -9.91.]

Therefore the equation of the tangent line to the implicitly defined function (sin y + y^3=6-x^3) at the point ((sqrt[3]{6},0)) is [y = -3sqrt[3]{36}(x-sqrt[3]{6})+0 approx -9.91x+18.]The curve and this tangent line are shown in Figure 2.20.

This suggests a general method for implicit differentiation. For the steps below assume (y) is a function of (x).

  1. Take the derivative of each term in the equation. Treat the (x) terms like normal. When taking the derivatives of (y) terms, the usual rules apply except that, because of the Chain Rule, we need to multiply each term by (y^prime ).
  2. Get all the (y^prime ) terms on one side of the equal sign and put the remaining terms on the other side.
  3. Factor out (y^prime ); solve for (y^prime ) by dividing.

Practical Note: When working by hand, it may be beneficial to use the symbol (frac{dy}{dx}) instead of (y^prime ), as the latter can be easily confused for (y) or (y^1).

Example 69: Using Implicit Differentiation

Given the implicitly defined function (y^3+x^2y^4=1+2x), find (y^prime ).

Solution

We will take the implicit derivatives term by term. The derivative of (y^3) is (3y^2y^prime ).

The second term, (x^2y^4), is a little tricky. It requires the Product Rule as it is the product of two functions of (x): (x^2) and (y^4). Its derivative is (x^2(4y^3y^prime ) + 2xy^4). The first part of this expression requires a (y^prime ) because we are taking the derivative of a (y) term. The second part does not require it because we are taking the derivative of (x^2).

The derivative of the right hand side is easily found to be (2). In all, we get:

[3y^2y^prime + 4x^2y^3y^prime + 2xy^4 = 2.]

Move terms around so that the left side consists only of the (y^prime ) terms and the right side consists of all the other terms:

[3y^2y^prime + 4x^2y^3y^prime = 2-2xy^4.]

Factor out (y^prime ) from the left side and solve to get

[y^prime = frac{2-2xy^4}{3y^2+4x^2y^3}.]

To confirm the validity of our work, let's find the equation of a tangent line to this function at a point. It is easy to confirm that the point ((0,1)) lies on the graph of this function. At this point, (y^prime = 2/3). So the equation of the tangent line is (y = 2/3(x-0)+1). The function and its tangent line are graphed in Figure 2.21.

Notice how our function looks much different than other functions we have seen. For one, it fails the vertical line test. Such functions are important in many areas of mathematics, so developing tools to deal with them is also important.

Example 70: Using Implicit Differentiation

Given the implicitly defined function (sin(x^2y^2)+y^3=x+y), find (y^prime ).

Solution

Differentiating term by term, we find the most difficulty in the first term. It requires both the Chain and Product Rules.

[egin{align*} frac{d}{dx}Big(sin(x^2y^2)Big) &= cos(x^2y^2)cdotfrac{d}{dx}Big(x^2y^2Big) &= cos(x^2y^2)cdotig(x^2(2yy^prime )+2xy^2ig) &= 2(x^2yy^prime +xy^2)cos(x^2y^2). end{align*} ]

We leave the derivatives of the other terms to the reader. After taking the derivatives of both sides, we have

[2(x^2yy^prime +xy^2)cos(x^2y^2) + 3y^2y^prime = 1 + y^prime .]

We now have to be careful to properly solve for (y^prime ), particularly because of the product on the left. It is best to multiply out the product. Doing this, we get

[2x^2ycos(x^2y^2)y^prime + 2xy^2cos(x^2y^2) + 3y^2y^prime = 1 + y^prime .]

From here we can safely move around terms to get the following:

[2x^2ycos(x^2y^2)y^prime + 3y^2y^prime - y^prime = 1 - 2xy^2cos(x^2y^2).]

Then we can solve for (y^prime ) to get

[y^prime = frac{1 - 2xy^2cos(x^2y^2)}{2x^2ycos(x^2y^2)+3y^2-1}.]

A graph of this implicit function is given in Figure 2.22. It is easy to verify that the points ((0,0)), ((0,1)) and ((0,-1)) all lie on the graph. We can find the slopes of the tangent lines at each of these points using our formula for (y^prime ).

At ((0,0)), the slope is (-1).

At ((0,1)), the slope is (1/2).

At ((0,-1)), the slope is also (1/2).

The tangent lines have been added to the graph of the function in Figure 2.23.

Quite a few "famous'' curves have equations that are given implicitly. We can use implicit differentiation to find the slope at various points on those curves. We investigate two such curves in the next examples.

Example 71: Finding slopes of tangent lines to a circle

Find the slope of the tangent line to the circle (x^2+y^2=1) at the point ((1/2, sqrt{3}/2)).

Solution

Taking derivatives, we get (2x+2yy^prime =0). Solving for (y^prime ) gives: [ y^prime = frac{-x}{y}.]

This is a clever formula. Recall that the slope of the line through the origin and the point ((x,y)) on the circle will be (y/x). We have found that the slope of the tangent line to the circle at that point is the opposite reciprocal of (y/x), namely, (-x/y). Hence these two lines are always perpendicular.

At the point ((1/2, sqrt{3}/2)), we have the tangent line's slope as

[y^prime = frac{-1/2}{sqrt{3}/2} = frac{-1}{sqrt{3}} approx -0.577.]

A graph of the circle and its tangent line at ((1/2,sqrt{3}/2)) is given in Figure 2.24, along with a thin dashed line from the origin that is perpendicular to the tangent line. (It turns out that all normal lines to a circle pass through the center of the circle.)

This section has shown how to find the derivatives of implicitly defined functions, whose graphs include a wide variety of interesting and unusual shapes. Implicit differentiation can also be used to further our understanding of "regular'' differentiation.

One hole in our current understanding of derivatives is this: what is the derivative of the square root function? That is, [frac{d}{dx}ig(sqrt{x}ig) = frac{d}{dx}ig(x^{1/2}ig) = ext{?}]

We allude to a possible solution, as we can write the square root function as a power function with a rational (or, fractional) power. We are then tempted to apply the Power Rule and obtain [frac{d}{dx}ig(x^{1/2}ig) = frac12x^{-1/2} = frac{1}{2sqrt{x}}.]

The trouble with this is that the Power Rule was initially defined only for positive integer powers, (n>0). While we did not justify this at the time, generally the Power Rule is proved using something called the Binomial Theorem, which deals only with positive integers. The Quotient Rule allowed us to extend the Power Rule to negative integer powers. Implicit Differentiation allows us to extend the Power Rule to rational powers, as shown below.

Let (y = x^{m/n}), where (m) and (n) are integers with no common factors (so (m=2) and (n=5) is fine, but (m=2) and (n=4) is not). We can rewrite this explicit function implicitly as (y^n = x^m). Now apply implicit differentiation.

[egin{align*}y &= x^{m/n} y^n &= x^m frac{d}{dx}ig(y^nig) &= frac{d}{dx}ig(x^mig) ncdot y^{n-1}cdot y^prime &= mcdot x^{m-1} y^prime &= frac{m}{n} frac{x^{m-1}}{y^{n-1}} quad ext{(now substitute (x^{m/n}) for (y))} &= frac{m}{n} frac{x^{m-1}}{(x^{m/n})^{n-1}} quad ext{(apply lots of algebra)} &= frac{m}n x^{(m-n)/n} &= frac{m}n x^{m/n -1}.end{align*}]

The above derivation is the key to the proof extending the Power Rule to rational powers. Using limits, we can extend this once more to include all powers, including irrational (even transcendental!) powers, giving the following theorem.

Theorem 21: Power Rule for Differentiation

Let (f(x) = x^n), where (n eq 0) is a real number. Then (f) is a differentiable function, and (f^prime(x) = ncdot x^{n-1}).

This theorem allows us to say the derivative of (x^pi) is (pi x^{pi -1}).

We now apply this final version of the Power Rule in the next example, the second investigation of a "famous'' curve.

Example 72: Using the Power Rule

Find the slope of (x^{2/3}+y^{2/3}=8) at the point ((8,8)).

Solution

This is a particularly interesting curve called an astroid. It is the shape traced out by a point on the edge of a circle that is rolling around inside of a larger circle, as shown in Figure 2.25.

To find the slope of the astroid at the point ((8,8)), we take the derivative implicitly.

[egin{align*} frac23x^{-1/3}+frac23y^{-1/3}y^prime &=0 frac23y^{-1/3}y^prime &= -frac23x^{-1/3} y^prime &= -frac{x^{-1/3}}{y^{-1/3}} y^prime &= -frac{y^{1/3}}{x^{1/3}} = -sqrt[3]{frac{y}x}. end{align*}]

Plugging in (x=8) and (y=8), we get a slope of (-1). The astroid, with its tangent line at ((8,8)), is shown in Figure 2.26.

Implicit Differentiation and the Second Derivative

We can use implicit differentiation to find higher order derivatives. In theory, this is simple: first find (frac{dy}{dx}), then take its derivative with respect to (x). In practice, it is not hard, but it often requires a bit of algebra. We demonstrate this in an example.

Example 73: Finding the second derivative

Given (x^2+y^2=1), find (frac{d^2y}{dx^2} = y^{primeprime}).

Solution

We found that (y^prime = frac{dy}{dx} = -x/y) in Example 71. To find (y^{primeprime}), we apply implicit differentiation to (y^prime ).

[egin{align*} y^{primeprime} &= frac{d}{dx}ig(y^prime ig) &= frac{d}{dx}left(-frac xy ight)qquad ext{(Now use the Quotient Rule.)} &= -frac{y(1) - x(y^prime )}{y^2} end{align*}]

replace (y^prime ) with (-x/y):

[egin{align*}&= -frac{y-x(-x/y)}{y^2} &= -frac{y+x^2/y}{y^2}.end{align*}]

While this is not a particularly simple expression, it is usable. We can see that (y^{primeprime}>0) when (y<0) and (y^{primeprime}<0) when (y>0). In Section 3.4, we will see how this relates to the shape of the graph.

Logarithmic Differentiation

Consider the function (y=x^x); it is graphed in Figure 2.27. It is well--defined for (x>0) and we might be interested in finding equations of lines tangent and normal to its graph. How do we take its derivative?

The function is not a power function: it has a "power'' of (x), not a constant. It is not an exponential function: it has a "base'' of (x), not a constant.

A differentiation technique known as logarithmic differentiation becomes useful here. The basic principle is this: take the natural log of both sides of an equation (y=f(x)), then use implicit differentiation to find (y^prime ). We demonstrate this in the following example.

Example 74: Using Logarithmic Differentiation

Given (y=x^x), use logarithmic differentiation to find (y^prime ).

Solution

As suggested above, we start by taking the natural log of both sides then applying implicit differentiation.

[egin{align*} y &= x^x ln (y) &= ln (x^x) ext{(apply logarithm rule)} ln (y) &= xln x ext{(now use implicit differentiation)} frac{d}{dx}Big(ln (y)Big) &= frac{d}{dx}Big(xln xBig) frac{y^prime }{y} &= ln x + xcdotfrac1x frac{y^prime }{y} &= ln x + 1 y^prime &= yig(ln x+1ig) ext{(substitute (y=x^x))} y^prime &= x^xig(ln x+1ig). end{align*} ]

To "test'' our answer, let's use it to find the equation of the tangent line at (x=1.5). The point on the graph our tangent line must pass through is ((1.5, 1.5^{1.5}) approx (1.5, 1.837)). Using the equation for (y^prime ), we find the slope as

[y^prime = 1.5^{1.5}ig(ln 1.5+1ig) approx 1.837(1.405) approx 2.582.]

Thus the equation of the tangent line is (y = 1.6833(x-1.5)+1.837). Figure 2.28 graphs (y=x^x) along with this tangent line.

Implicit differentiation proves to be useful as it allows us to find the instantaneous rates of change of a variety of functions. In particular, it extended the Power Rule to rational exponents, which we then extended to all real numbers. In the next section, implicit differentiation will be used to find the derivatives of inverse functions, such as (y=sin^{-1} x).


Sometimes the implicit way works where the explicit way is hard or impossible.

Example: 10x 4 - 18xy 2 + 10y 3 = 48

How do we solve for y? We don't have to!

  • First, differentiate with respect to x (use the Product Rule for the xy 2 term).
  • Then move all dy/dx terms to the left side.
  • Solve for dy/dx

(the middle term is explained below)

dy dx = 9y 2 − 20x 3
3(5y 2 − 6xy)

Product Rule

For the middle term we used the Product Rule: (fg)’ = f g’ + f’ g

Because (y 2 )’ = 2y dy dx (we worked that out in a previous example)

Oh, and dxdx = 1, in other words x’ = 1


SECTION 3.5 Implicit Differentiation 215 3.5 EXERCISES 1-4 29. xy (2x 2y2 x, (0.). (cardioid) (a) Find y' by implicit differentiation. (b) Solve the equation explicitly for y and differentiate to get y in terms of x. (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for y into your solution for part (a). 1. 9x2 y2 1 2. 2x2x + xy 1 3. 1 2 4. 30. x2/y2/4, (-3/3, 1), (astroid) - =4 5-20 Find dy/dx by implicit differentiation. 5. x2-4xy + y2 4 6. 2x2+xy y2 = 2 7. x+x'y2y 5 8. x-xy2 y= 1 31. 2(x2y225(x2- y2), ( 3, 1), (lemniscate) x2 10. xe xy 9. = y2 + 1 x+ y 12. Cos(xy) 1 + sin y 11. y cos x x2 + y 14. e' sin x x + xy 13. Vx +y x+ y Vx2 +y 15. e xy 16. xy 32. y(y 4) x(x -5), (0,-2), (devil's curve) 17. tan (xy) x + xy 18. x sin y + y sin x 1 20. tan(x y) 19. sin(xy) cos(x + y) 1x2 21. If f(x) +x[f (x)]' = 10 and f(1) = 2, find f'(1). 22. If g(x) +x sin g(x) x, find g' (0). 33. (a) The curve with equationy 5x-x2 is called a kampyle of Eudoxus. Find an equation of the tam line to this curve at the point (1, 2). (b) Illustrate part (a) by graphing the curve and the t line on a common screen. (If your graphing devi- graph implicitly defined curves, then use that ca ity. If not, you can still graph this curve by grap upper and lower halves separately.) 23-24 Regard y as the independent variable and x as the depen- dent variable and use implicit differentiation to find dx/dy. 24. y sec x x tan y 23. xy2-x'y + 2xy 0 25-32 Use implicit differentiation to find an equation of the tangent line to the curve at the given point. (T/2, T/4) x +3x2 is calle 34. (a) The curve with equation y Tschirnhausen cubic. Find an equation of the line to this curve at the point (1, -2). (b) At what points does this curve have horizonta tangents? (c) Illustrate parts (a) and (b) by graphing the cu tangent lines on a common screen. 25. y sin 2x x cos 2y, (T, T) 2x 2y, 26. sin(x + y) 27. xxy y2= 1, (2, 1) (hyperbola) 28. x2xy +4y 12, (2, 1) (ellipse)

I need help finding dy/dx for question #9 in Section 3.5, page 215, of the James Stewart Calculus Eighth Edition textbook.

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SECTION 3.5 Implicit Differentiation 215 3.5 EXERCISES 1-4 29. xy (2x 2y2 x, (0.). (cardioid) (a) Find y' by implicit differentiation. (b) Solve the equation explicitly for y and differentiate to get y in terms of x. (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for y into your solution for part (a). 1. 9x2 y2 1 2. 2x2x + xy 1 3. 1 2 4. 30. x2/y2/4, (-3/3, 1), (astroid) - =4 5-20 Find dy/dx by implicit differentiation. 5. x2-4xy + y2 4 6. 2x2+xy y2 = 2 7. x+x'y2y 5 8. x-xy2 y= 1 31. 2(x2y225(x2- y2), ( 3, 1), (lemniscate) x2 10. xe xy 9. = y2 + 1 x+ y 12. Cos(xy) 1 + sin y 11. y cos x x2 + y 14. e' sin x x + xy 13. Vx +y x+ y Vx2 +y 15. e xy 16. xy 32. y(y 4) x(x -5), (0,-2), (devil's curve) 17. tan (xy) x + xy 18. x sin y + y sin x 1 20. tan(x y) 19. sin(xy) cos(x + y) 1x2 21. If f(x) +x[f (x)]' = 10 and f(1) = 2, find f'(1). 22. If g(x) +x sin g(x) x, find g' (0). 33. (a) The curve with equationy 5x-x2 is called a kampyle of Eudoxus. Find an equation of the tam line to this curve at the point (1, 2). (b) Illustrate part (a) by graphing the curve and the t line on a common screen. (If your graphing devi- graph implicitly defined curves, then use that ca ity. If not, you can still graph this curve by grap upper and lower halves separately.) 23-24 Regard y as the independent variable and x as the depen- dent variable and use implicit differentiation to find dx/dy. 24. y sec x x tan y 23. xy2-x'y + 2xy 0 25-32 Use implicit differentiation to find an equation of the tangent line to the curve at the given point. (T/2, T/4) x +3x2 is calle 34. (a) The curve with equation y Tschirnhausen cubic. Find an equation of the line to this curve at the point (1, -2). (b) At what points does this curve have horizonta tangents? (c) Illustrate parts (a) and (b) by graphing the cu tangent lines on a common screen. 25. y sin 2x x cos 2y, (T, T) 2x 2y, 26. sin(x + y) 27. xxy y2= 1, (2, 1) (hyperbola) 28. x2xy +4y 12, (2, 1) (ellipse)


Math 140 Calculus with Analytic Geometry I

Blue Book Description: Calculus is an important building block in the education of any professional who uses quantitative analysis. This course introduces and develops the mathematical skills required for analyzing change and creating mathematical models that replicate real-life phenomena. The goals of our calculus courses include to develop the students’ knowledge of calculus techniques and to use the calculus environment to develop critical thinking and problem solving skills. The concept of limit is central to calculus MATH 140 begins with a study of this concept. Differential calculus topics include derivatives and their applications to rates of change, related rates, linearization, optimization, and graphing techniques. The Fundamental Theorem of Calculus, relating differential and integral calculus begins the study of Integral Calculus. Antidifferentiation and the technique of substitution is used in integration applications of finding areas of plane figures and volumes of solids of revolution. Trigonometric functions are included in every topic. Students may only take one course for credit from MATH 110, 140, 140A, 140B, and 140H.

Pre-requisites: Math 22 and Math 26 or Math 26 and satisfactory performance on the mathematics placement examination or Math 40 or Math 41 or satisfactory performance on the mathematics placement examination.

Pre-requisite for: MATH 141, MATH 141B, MATH 220

Bachelor of Arts: Quantification
General Education: Quantification (GQ)
GenEd Learning Objective: Crit and Analytical Think
GenEd Learning Objective: Key Literacies

Suggested Textbook:
Single Variable Calculus: Early Transcendentals, Vol 1, 8th edition, by James Stewart, published by Brookes/Cole Cengage Learning.
Check with your instructor to make sure this is the textbook used for your section.

Topics:
Chapter 2: Limits and Derivatives
2.1 The Tangent and Velocity Problems
2.2 The Limit of a Function
2.3 Calculating Limits Using the Limit Laws
2.4 The Precise Definition of a Limit (optional),
2.5 Continuity
2.6 Limits at Infinity Horizontal Asymptotes
2.7 Derivatives and Rates of Change
2.8 The Derivative as a Function

Chapter 3: Differentiation Rules
3.1 Derivatives of Polynomials and Exponential Functions
3.2 The Product and Quotient Rules
3.3 Derivatives of Trigonometric Functions
3.4 The Chain Rule
3.5 Implicit Differentiation
3.6 Derivatives of Logarithmic Functions
3.7 Rates of Change in Natural and Social Sciences (optional)
3.8 Exponential Growth and Decay (optional)
3.9 Related Rates

Chapter 4: Applications of Differentiation
4.1 Maximum and Minimum Values
4.2 The Mean Value Theorem
4.3 How Derivatives Affect the Shape of a Graph
4.4 Indeterminate Forms and L’Hospital’s Rule
4.5 Summary of Curve Sketching
4.7 Optimization Problems
4.8 Newton’s Method (optional)
4.9 Antiderivatives

Chapter 5: Integrals
5.1 Areas and Distance
5.2 The Definite Integral
5.3 The Fundamental Theorem of Calculus
5.4 Indefinite Integrals and the Net Change Theorem
5.5 The Substitution Rule

Chapter 6: Applications of Integration
6.1 Areas between Curve
6.2 Volumes
6.2 Volumes by Cylindrical Shells (optional)
6.4 Work (optional)
6.5 Average Value of a Function (optional)


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Last reply by: Arshin Jain
Sun Dec 15, 2013 11:22 PM

Post by Steve Denton on October 11, 2012

On ex.5 at 16:00, adding the numerator terms requires multiplying the first term by sin y/sin y and therefore should be sin x sin^2 y. Right?

Last reply by: Steve Denton
Thu Oct 11, 2012 7:59 PM

Post by James Xie on July 5, 2012

For the problem: x + sin(x) = (x^2)(y^2), this is the work I did:
1) 1 + cos(y)(dy/dx) = 2(x^2)y(dy/dx) + 2x(y^2)
2)(dy/dx)(cos(y)-2(x^2)y) = 2x(y^2)-1
3)(dy/dx) = (2x(y^2)-1)/(cos(y)-2(x^2)y)
Would this be correct too? (Sorry if it looks unclear here)

Last reply by: Joshua Spears
Tue Feb 26, 2013 1:06 PM

Post by James Xie on July 5, 2012

For the second order example, why is it 4(8y+1)^2 in the beginning of the final answer? Shouldn't it start with 4(8y+1)?


Implicit Differentiation

which are the top and bottom halves of a cricle respectively, to define the functional relation completely.

And xy = sin( y )+ x 2 y 2 (the magenta curves in the figure at the left) cannot be solved for either y as an explicit function of x or x as an explicit function of y . This implicit function is considered in Example 2.

Perhaps surprisingly, we can take the derivative of implicit functions just as we take the derivative of explicit functions. We simply take the derivative of each side of the equation, remembering to treat the dependent variable as a function of the independent variable, apply the rules of differentiation, and solve for the derivative. Returning to our original example:

This is of course what we get from differentiating the explicit form, y = 2 x -3, with respect to x .

This simple example may not be very enlightening. Consider the second example, the equation that describes a circle of radius 3, centered at the origin. Taking the derivative of both sides with respect to x , using the power rule for the derivative of y ,

It can be seen from the figures that for either part of the circle, the slope of the tangent line has the opposite sign of the ratio x / y , and that the magnitude of the slope becomes larger as the tangent point nears the x -axis.

(For this case, finding dy/dx as an explicit function of x requires use of the power rule for fractional powers, usually considered later. This example may be thought of as a taste of things to come.)

Some examples:

Note that this expression can be solved to give x as an explicit function of y by solving a cubic equation, and finding y as an explicit function of x would involve soving a quartic equation, neither of which is in our plan.

Using the chain rule and treating y as an implicit function of x ,

In this case, the chain rule and product rule are both used to advantage:

The use of inverse trigonometric functions allow this to be solve for y as an explicit function of x and graphed, as shown. However, this function serves as a good example of implicit differentiation:


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Math 144: Business Calculus

1) 8/22/2017 - Go over syllabus. Review some important concepts from College Algebra.
Algebra Review handout
TUTORIAL VIDEO (Imporant): ALGEBRA REVIEW-Fractions, Exponents, Inequalities, Solving Equations.
Tutorial video: What is a function?


2) 8/24/2017- A little bit more review in the beginning of class then start on Chapter 2.1- Introduction to limit.
Tutorial video: Rational expressions and domain
Tutorial video: Why Limits are Important in Calculus
Tutorial video: Introduction to limit

3) 8/29/2017- Chapter 2.1 continue. Properties of limits, limits of polynomial and rational functions, intermediate form.
****HOMEWORK 1 IS DUE TONIGHT @ 11:59 PM.
Tutorial video: Properties of limits
Tutorial video: limits of polynomial functions
Tutorial video: limits of polynomial and rational functions

4)8/31/2017 - Chapter 2.1 Cont. ( Yes, we spent 3 class periods on this section. This is a very important concept and I want you to fully understand it).
**Tutorial video:** One-sided limits from graphs- Related to Problem 4-7 on HW2.


10) 9/26/2017- Chapter 3.1 - The constant "e". Review of exponential and logarithm function. Define the value "e" in term of limit.
Tutorial video: The constant "e" as a limit
Tutorial video: An Introduction to Exponential Functions
Tutorial video: Graphing exponential functions

12)10/03/2017 - Chapter 3.2 - Derivative of exponential and logarithmic functions.
We did 8 different examples in the beginning of class.

Chapter 3.3 - Product rule. We will do the quotient rule on Thursday 10/5/2017.
Tutorial video: Examples of product rule
Tutorial video: More examples of product rule
Tutorial video: Proof of product rule (In case you are intersted)

15) 10/10/2017 - Chapter 3.4- "Chain Rule" continue.
Tutorial video: More examples of chain rule.
Please IGNORE the part about TRIG functions. We don't study trig function in this class. Tutorial video: Another example of chain rule
Tutorial video: More examples of chain rule
Tutorial video: Even more examples chain rule :-)

16) 10/12/2017 - Chapter 3.5- Implicit differentiation Cont.
Tutorial video: Examples of implicit differentiation
Tutorial video: More examples but PLEASE IGNOTE TRIG FUNCTIONS (Sin(x) Cos(x) etc.) since we do not study trig functions in this course
logarithm differentiation using implicit differentiation https://www.youtube.com/watch?v=Q27MGfI1V70

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MATH 161: Calculus I

Note: MATH 162A uses a different textbook. Namely, James Stewart. Calculus, Early Transcendentals (WebAssign eBook) 8th ed. Cengage Learning. Be sure you are reading the correct information.

Chapter 1: Functions and Models
1.1 Four Ways to Represent a Function
1.2 Mathematical Models: A Catalog of Essential Functions
1.3 New Functions from Old Functions
1.4 Exponential Functions and Logarithms
1.5 Inverse Functions and Logarithms
Optional: Graphing with calculators, Mathematica, Wolfram Alpha (pp. xxiv-xxv)

Chapter 2: Limits and Derivatives
2.1 The Tangent and Velocity Problems
2.2 The Limit of a Function
2.3 Calculating Limits Using the Limit Laws
2.4 The Precise Definition of a Limit
2.5 Continuity
2.6 Limits at Infinity Horizontal Asymptotes
2.7 Derivatives and Rates of Change
2.8 The Derivative as a Function

Chapter 3: Differentiation Rules
3.1 Derivatives of Polynomials and Exponential Functions
3.2 The Product and Quotient Rules
3.3 Derivatives of Trigonometric Functions
3.4 The Chain Rule
3.5 Implicit Differentiation
3.6 Derivatives of Logarithmic Functions
3.7 Rates of Change in Natural and Social Sciences
3.8 Exponential Growth and Decay
3.9 Related Rates
3.10 Linear Approximations and Differentials
3.11 Optional: Hyperbolic Functions

Chapter 4: Applications of Derivatives
4.1 Maximum and Minimum Values
4.2 The Mean Value Theorem
4.3 How Derivatives Affect the Shape of a Graph
4.4 Indeterminate Forms and l'Hospital's Rule
4.5 Summary of Curve Sketching
4.6 Optional: Graphing with Calculus and Calculators
4.7 Optimization Problems
4.8 Optional: Newton's Method
4.9 Antiderivatives

Chapter 5: Integrals
5.1 Areas and Distances
5.2 The Definite Integral
5.3 The Fundamental Theorem of Calculus
5.4 Indefinite Integrals and the Net Change Theorem
5.5 The Substitution Rule


Watch the video: Derivative of sin3x5 Calculus 1 Includes Chain Rule (December 2021).