IMPLICIT DIFFERENTIATION PROBLEMS

The following problems require the use of implicit differentiation. Implicit differentiation is nothing more than a special case of the well-known chain rule for derivatives. The majority of differentiation problems in first-year calculus involve functions y written EXPLICITLY as functions of x . For example, if

$y = 3x^2 -\sin(7x+5)$ ,

then the derivative of y is

$y' = 6x - 7 \cos(7x+5)$ .

However, some functions y are written IMPLICITLY as functions of x . A familiar example of this is the equation

x² + y² = 25 ,

which represents a circle of radius five centered at the origin. Suppose that we wish to find the slope of the line tangent to the graph of this equation at the point (3, -4) .

How could we find the derivative of y in this instance ? One way is to first write y explicitly as a function of x . Thus,

x² + y² = 25 ,

y² = 25 - x² ,

and

$y = \pm \sqrt{ 25 - x^2 }$ ,

where the positive square root represents the top semi-circle and the negative square root represents the bottom semi-circle. Since the point (3, -4) lies on the bottom semi-circle given by

$y = - \sqrt{ 25 - x^2 }$ ,

the derivative of y is

$y' = - (1/2) \big( 25 - x^2 \big)^{-1/2} (-2x) = \displaystyle{ x \over \sqrt{ 25 - x^2 } }$ ,

i.e.,

$y' = \displaystyle{ x \over \sqrt{ 25 - x^2 } }$ .

Thus, the slope of the line tangent to the graph at the point (3, -4) is

$m = y' = \displaystyle{ (3) \over \sqrt{ 25 - (3)^2 } } = \displaystyle{ 3 \over 4 }$ .

Unfortunately, not every equation involving x and y can be solved explicitly for y . For the sake of illustration we will find the derivative of y WITHOUT writing y explicitly as a function of x . Recall that the derivative (D) of a function of x squared, (f(x))² , can be found using the chain rule :

$D \{ ( f(x) )^2 \} = 2 f(x) \ D \{ f(x) \} = 2 f(x) f'(x)$ .

Since y symbolically represents a function of x, the derivative of y² can be found in the same fashion :

$D \{ y^2 \} = 2 y \ D \{ y \} = 2 y y'$ .

Now begin with

x² + y² = 25 .

Differentiate both sides of the equation, getting

D ( x² + y² ) = D ( 25 ) ,

D ( x² ) + D ( y² ) = D ( 25 ) ,

and

2x + 2 y y' = 0 ,

so that

2 y y' = - 2x ,

and

$y' = \displaystyle{ - 2x \over 2y } = \displaystyle{ - x \over y }$ ,

i.e.,

$y' = \displaystyle{ - x \over y }$ .

Thus, the slope of the line tangent to the graph at the point (3, -4) is

$m = y' = \displaystyle{ - (3) \over (-4) } = \displaystyle{ 3 \over 4 }$ .

This second method illustrates the process of implicit differentiation. It is important to note that the derivative expression for explicit differentiation involves x only, while the derivative expression for implicit differentiation may involve BOTH x AND y .

The following problems range in difficulty from average to challenging.

• PROBLEM 1 : Assume that y is a function of x . Find y' = dy/dx for x³ + y³ = 4 .

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• PROBLEM 2 : Assume that y is a function of x . Find y' = dy/dx for (x-y)² = x + y - 1 .

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• PROBLEM 3 : Assume that y is a function of x . Find y' = dy/dx for $y = \sin(3x + 4y)$ .

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• PROBLEM 4 : Assume that y is a function of x . Find y' = dy/dx for y = x² y³ + x³ y² .

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• PROBLEM 5 : Assume that y is a function of x . Find y' = dy/dx for e^xy = e^4x - e^5y .

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• PROBLEM 6 : Assume that y is a function of x . Find y' = dy/dx for $\cos^2 x + \cos^2 y = \cos( 2x + 2y )$ .

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• PROBLEM 7 : Assume that y is a function of x . Find y' = dy/dx for x=3 + sqrt{x^2+y^2} .

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• PROBLEM 8 : Assume that y is a function of x . Find y' = dy/dx for $${ x - y^3 \over y + x^2 } = x + 2$$ .

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• PROBLEM 9 : Assume that y is a function of x . Find y' = dy/dx for $${ { y \over x^3 } + { x \over y^3 } } = x^2y^4$$ .

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• PROBLEM 10 : Find an equation of the line tangent to the graph of (x²+y²)³ = 8x²y² at the point (-1, 1) .

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• PROBLEM 11 : Find an equation of the line tangent to the graph of x² + (y-x)³ = 9 at x=1 .

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• PROBLEM 12 : Find the slope and concavity of the graph of x²y + y⁴ = 4 + 2x at the point (-1, 1) .

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• PROBLEM 13 : Consider the equation x² + xy + y² = 1 . Find equations for y' and y'' in terms of x and y only.

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• PROBLEM 14 : Find all points (x, y) on the graph of x^2/3 + y^2/3 = 8 (See diagram.) where lines tangent to the graph at (x, y) have slope -1 .

  

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• PROBLEM 15 : The graph of x² - xy + y² = 3 is a "tilted" ellipse (See diagram.). Among all points (x, y) on this graph, find the largest and smallest values of y . Among all points (x, y) on this graph, find the largest and smallest values of x .

  

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• PROBLEM 16 : Find all points (x, y) on the graph of (x²+y²)² = 2x²-2y² (See diagram.) where y' = 0.

  

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Click HERE to return to the original list of various types of calculus problems.

Your comments and suggestions are welcome. Please e-mail any correspondence to Duane Kouba by clicking on the following address :

kouba@math.ucdavis.edu

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