how to find dy/dx

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

x ² + y ² = 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,

x ² + y ² = 25 ,

y ² = 25 - x ² ,

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 y ² can be found in the same fashion :

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

Now begin with

x ² + y ² = 25 .

Differentiate both sides of the equation, getting

D ( x ² + y ² ) = D ( 25 ) ,

D ( x ² ) + D ( y ² ) = 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 x ³ + y ³ = 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 = x ² y ³ + x ³ y ² .

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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 $\displaystyle{ x - y^3 \over y + x^2 } = x + 2$ .

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

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PROBLEM 10 : Find an equation of the line tangent to the graph of (x ²+y ²)³ = 8x ² y ² at the point (-1, 1) .

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PROBLEM 11 : Find an equation of the line tangent to the graph of x ² + (y-x)³ = 9 at x=1 .

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

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PROBLEM 13 : Consider the equation x ² + xy + y ² = 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 x ² - xy + y ² = 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 (x ²+y ²)² = 2x ²-2y ² (See diagram.) where y' = 0.

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