Differentiationap
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. Alternatively, one can use implicit differentiation a second time to get Substitutmg x = O, y 3, and y' — gives. 1982 Solution (b) = 152 Implicit: 2X—+. If 5x^2+5x+xy=2 and y(2)= -14 find y'(2) by implicit differentiation. Use implicit differentiation to find an equation of the tangent line to the curve at the given point. X2 + 2xy − y2 + x = 17, (3, 5) (hyperbola) Math. Suppose that x and y are related by the equation x^2/4 + y^3/2 = 4.

SOLUTIONS TO IMPLICIT DIFFERENTIATION PROBLEMS


Implicit Differentiation. Finding the derivative when you can’t solve for y. You may like to read Introduction to Derivatives and Derivative Rules first. Implicit vs Explicit. A function can be explicit or implicit: Explicit: 'y = some function of x'. When we know x we can calculate y directly. Implicit: 'some function of y and x equals.
SOLUTION 1 : Begin with x3 + y3 = 4 . Differentiate both sides of the equation, getting

D ( x3 + y3 ) = D ( 4 ) ,

D ( x3 ) + D ( y3 ) = D ( 4 ) ,

(Remember to use the chain rule on D ( y3 ) .)

3x2 + 3y2y' = 0 ,

so that (Now solve for y' .)

3y2y' = - 3x2 ,

and

.

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SOLUTION 2 : Begin with (x-y)2 = x + y - 1 . Differentiate both sides of the equation, getting

D (x-y)2 = D ( x + y - 1 ) ,

D (x-y)2 = D ( x ) + D ( y ) - D ( 1 ) ,

(Remember to use the chain rule on D (x-y)2 .)

,

2 (x-y) (1- y') = 1 + y' ,

so that (Now solve for y' .)

2 (x-y) - 2 (x-y) y' = 1 + y' ,

- 2 (x-y) y' - y' = 1 - 2 (x-y) ,

(Factor out y' .)

y' [ - 2 (x-y) - 1 ] = 1 - 2 (x-y) ,

and

.

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SOLUTION 3 : Begin with . Differentiate both sides of the equation, getting

4.1 Implicit Differentiationap Calculus Solver

,

(Remember to use the chain rule on .)

,

,

so that (Now solve for y' .)

,

,

(Factor out y' .)

,

and

.

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SOLUTION 4 : Begin with y = x2y3 + x3y2 . Differentiate both sides of the equation, getting

D(y) = D ( x2y3 + x3y2 ) ,

D(y) = D ( x2y3 ) + D ( x3y2 ) ,

(Use the product rule twice.)

,

(Remember to use the chain rule on D ( y3 ) and D ( y2 ) .)

,

y' = 3x2y2y' + 2x y3 + 2x3y y' + 3x2y2 ,

so that (Now solve for y' .)

y' - 3x2y2y' - 2x3y y' = 2x y3 + 3x2y2 ,

(Factor out y' .)

y' [ 1 - 3x2y2 - 2x3y ] = 2x y3 + 3x2y2 ,

and

.

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SOLUTION 5 : Begin with . Differentiate both sides of the equation, getting

,

,

,

,

so that (Now solve for .)

,

,

(Factor out .)

,

and

.

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SOLUTION 6 : Begin with . Differentiate both sides of the equation, getting

,

,

,

,

so that (Now solve for y' .)

,

,

(Factor out y' .)

,

,

,

and

.

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SOLUTION 7 : Begin with . Differentiate both sides of the equation, getting
4.1 implicit differentiationap calculus multiple choice

4.1 Implicit Differentiationap Calculus Multiple Choice

,

1 = (1/2)( x2 + y2 )-1/2D ( x2 + y2 ) ,

1 = (1/2)( x2 + y2 )-1/2 ( 2x + 2y y' ) ,

so that (Now solve for y' .)

,

,

,

,

and

.

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SOLUTION 8 : Begin with . Clear the fraction by multiplying both sides of the equation by y + x2 , getting

,

or

x - y3 = xy + 2y + x3 + 2x2 .

Now differentiate both sides of the equation, getting

D ( x - y3 ) = D ( xy + 2y + x3 + 2x2 ) ,

D ( x ) - D (y3 ) = D ( xy ) + D ( 2y ) + D ( x3 ) + D ( 2x2 ) ,

(Remember to use the chain rule on D (y3 ) .)

1 - 3 y2y' = ( xy' + (1)y ) + 2 y' + 3x2 + 4x ,

so that (Now solve for y' .)

1 - y - 3x2 - 4x = 3 y2y' + xy' + 2 y' ,

(Factor out y' .)

1 - y - 3x2 - 4x = (3y2 + x + 2) y' ,

and

.

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SOLUTION 9 : Begin with . Clear the fractions by multiplying both sides of the equation by x3y3 , getting

,

,

y4 + x4 = x5y7 .

Now differentiate both sides of the equation, getting

D ( y4 + x4 ) = D ( x5y7 ) ,

D ( y4 ) + D ( x4 ) = x5D (y7 ) + D ( x5 ) y7 ,

(Remember to use the chain rule on D (y4 ) and D (y7 ) .)

4.1 Implicit Differentiationap Calculus Calculator

4 y3y' + 4 x3 = x5 (7 y6y' ) + ( 5 x4 ) y7 ,

4.1 Implicit Differentiationap Calculus Algebra

so that (Now solve for y' .)

4 y3y' - 7 x5y6y' = 5 x4y7 - 4 x3 ,

(Factor out y' .)

y' [ 4 y3 - 7 x5y6 ] = 5 x4y7 - 4 x3 ,

and

.

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SOLUTION 10 : Begin with (x2+y2)3 = 8x2y2 . Now differentiate both sides of the equation, getting

D (x2+y2)3 = D ( 8x2y2 ) ,

3 (x2+y2)2D (x2+y2) = 8x2D (y2 ) + D ( 8x2 ) y2 ,

(Remember to use the chain rule on D (y2 ) .)

3 (x2+y2)2 ( 2x + 2 y y' ) = 8x2 (2 y y' ) + ( 16 x ) y2 ,

so that (Now solve for y' .)

6x (x2+y2)2 + 6 y (x2+y2)2y' = 16 x2y y' + 16 x y2 ,

6 y (x2+y2)2y' - 16 x2y y' = 16 x y2 - 6x (x2+y2)2 ,

(Factor out y' .)

y' [ 6 y (x2+y2)2 - 16 x2y ] = 16 x y2 - 6x (x2+y2)2 ,

and

.

Thus, the slope of the line tangent to the graph at the point (-1, 1) is

,

and the equation of the tangent line is

y - ( 1 ) = (1) ( x - ( -1 ) )

or

y = x + 2 .

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SOLUTION 11 : Begin with x2 + (y-x)3 = 9 . If x=1 , then

(1)2 + ( y-1 )3 = 9

so that

( y-1 )3 = 8 ,

y-1 = 2 ,

y = 3 ,

and the tangent line passes through the point (1, 3) . Now differentiate both sides of the original equation, getting

D ( x2 + (y-x)3 ) = D ( 9 ) ,

D ( x2 ) + D (y-x)3 = D ( 9 ) ,

2x + 3 (y-x)2D (y-x) = 0 ,

2x + 3 (y-x)2 (y'-1) = 0 ,

so that (Now solve for y' .)

2x + 3 (y-x)2y'- 3 (y-x)2 = 0 ,

3 (y-x)2y' = 3 (y-x)2 - 2x ,

and

.

Thus, the slope of the line tangent to the graph at (1, 3) is

,

and the equation of the tangent line is

y - ( 3 ) = (5/6) ( x - ( 1 ) ) ,

or

y = (5/6) x + (13/6) .

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SOLUTION 12 : Begin with x2y + y4 = 4 + 2x . Now differentiate both sides of the original equation, getting

D ( x2y + y4 ) = D ( 4 + 2x ) ,

D ( x2y ) + D (y4 ) = D ( 4 ) + D ( 2x ) ,

( x2y' + (2x) y ) + 4 y3y' = 0 + 2 ,

so that (Now solve for y' .)

x2y' + 4 y3y' = 2 - 2x y ,

(Factor out y' .)

y' [ x2 + 4 y3 ] = 2 - 2x y ,

and
(Equation 1)

.

Thus, the slope of the graph (the slope of the line tangent to the graph) at (-1, 1) is

.

Since y'= 4/5 , the slope of the graph is 4/5 and the graph is increasing at the point (-1, 1) . Now determine the concavity of the graph at (-1, 1) . Differentiate Equation 1, getting

.

Now let x=-1 , y=1 , and y'=4/5 so that the second derivative is

.

Since y' < 0 , the graph is concave down at the point (-1, 1) .

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Duane Kouba
1998-06-23