# 4.1 Implicit Differentiationap Calculus

Posted By admin On 29/12/21- 4.1 Implicit Differentiationap Calculus Multiple Choice
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For each problem, use implicit differentiation to find d2222y dx222 in terms of x and y. 13) 4y2 + 2 = 3x2 14) 5 = 4x2 + 5y2 Critical thinking question: 15) Use three strategies to find dy dx in terms of x and y, where 3x2 4y = x. Strategy 1: Use implicit differentiation directly on the given equation. Using Implicit Differentiation and the Product Rule. Assuming that y is defined implicitly by the equation x3siny + y = 4x + 3, find dy dx. D dx(x3siny + y) = d dx(4x + 3) Step 1: Differentiate both sides of the equation. D dx(x3siny) + d dx(y) = 4 Step 1.1: Apply the sum rule on the left.

### SOLUTIONS TO IMPLICIT DIFFERENTIATION PROBLEMS

* SOLUTION 1 :* Begin with *x*^{3} + *y*^{3} = 4 . Differentiate both sides of the equation, getting

*D* ( *x*^{3} + *y*^{3} ) = *D* ( 4 ) ,

*D* ( *x*^{3} ) + *D* ( *y*^{3} ) = *D* ( 4 ) ,

(Remember to use the chain rule on *D* ( *y*^{3} ) .)

3*x*^{2} + 3*y*^{2}*y*' = 0 ,

so that (Now solve for *y*' .)

3*y*^{2}*y*' = - 3*x*^{2} ,

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

,

(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* = *x*^{2}*y*^{3} + *x*^{3}*y*^{2} . Differentiate both sides of the equation, getting

*D*(*y*) = *D* ( *x*^{2}*y*^{3} + *x*^{3}*y*^{2} ) ,

*D*(*y*) = *D* ( *x*^{2}*y*^{3} ) + *D* ( *x*^{3}*y*^{2} ) ,

(Use the product rule twice.)

,

(Remember to use the chain rule on *D* ( *y*^{3} ) and *D* ( *y*^{2} ) .)

,

*y*' = 3*x*^{2}*y*^{2}*y*' + 2*x y*^{3} + 2*x*^{3}*y y*' + 3*x*^{2}*y*^{2} ,

so that (Now solve for *y*' .)

*y*' - 3*x*^{2}*y*^{2}*y*' - 2*x*^{3}*y y*' = 2*x y*^{3} + 3*x*^{2}*y*^{2} ,

(Factor out *y*' .)

*y*' [ 1 - 3*x*^{2}*y*^{2} - 2*x*^{3}*y* ] = 2*x y*^{3} + 3*x*^{2}*y*^{2} ,

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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### 4.1 Implicit Differentiationap Calculus Multiple Choice

*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

,

1 = (1/2)( *x*^{2} + *y*^{2} )^{-1/2}*D* ( *x*^{2} + *y*^{2} ) ,

1 = (1/2)( *x*^{2} + *y*^{2} )^{-1/2} ( 2*x* + 2*y 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* + *x*^{2} , getting

,

or

*x* - *y*^{3} = *xy* + 2*y* + *x*^{3} + 2*x*^{2} .

Now differentiate both sides of the equation, getting

*D* ( *x* - *y*^{3} ) = *D* ( *xy* + 2*y* + *x*^{3} + 2*x*^{2} ) ,

*D* ( *x* ) - *D* (*y*^{3} ) = *D* ( *xy* ) + *D* ( 2*y* ) + *D* ( *x*^{3} ) + *D* ( 2*x*^{2} ) ,

(Remember to use the chain rule on *D* (*y*^{3} ) .)

1 - 3 *y*^{2}*y*' = ( *xy*' + (1)*y* ) + 2 *y*' + 3*x*^{2} + 4*x* ,

so that (Now solve for *y*' .)

1 - *y* - 3*x*^{2} - 4*x* = 3 *y*^{2}*y*' + *xy*' + 2 *y*' ,

(Factor out *y*' .)

1 - *y* - 3*x*^{2} - 4*x* = (3*y*^{2} + *x* + 2) *y*' ,

and

.

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*SOLUTION 9 :* Begin with . Clear the fractions by multiplying both sides of the equation by *x*^{3}*y*^{3} , getting

,

,

*y*^{4} + *x*^{4} = *x*^{5}*y*^{7} .

Now differentiate both sides of the equation, getting

*D* ( *y*^{4} + *x*^{4} ) = *D* ( *x*^{5}*y*^{7} ) ,

*D* ( *y*^{4} ) + *D* ( *x*^{4} ) = *x*^{5}*D* (*y*^{7} ) + *D* ( *x*^{5} ) *y*^{7} ,

(Remember to use the chain rule on *D* (*y*^{4} ) and *D* (*y*^{7} ) .)

4 *y*^{3}*y*' + 4 *x*^{3} = *x*^{5} (7 *y*^{6}*y*' ) + ( 5 *x*^{4} ) *y*^{7} ,

so that (Now solve for *y*' .)

4 *y*^{3}*y*' - 7 *x*^{5}*y*^{6}*y*' = 5 *x*^{4}*y*^{7} - 4 *x*^{3} ,

(Factor out *y*' .)

*y*' [ 4 *y*^{3} - 7 *x*^{5}*y*^{6} ] = 5 *x*^{4}*y*^{7} - 4 *x*^{3} ,

and

.

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*SOLUTION 10 :* Begin with (*x*^{2}+*y*^{2})^{3} = 8*x*^{2}*y*^{2} . Now differentiate both sides of the equation, getting

*D* (*x*^{2}+*y*^{2})^{3} = *D* ( 8*x*^{2}*y*^{2} ) ,

3 (*x*^{2}+*y*^{2})^{2}*D* (*x*^{2}+*y*^{2}) = 8*x*^{2}*D* (*y*^{2} ) + *D* ( 8*x*^{2} ) *y*^{2} ,

(Remember to use the chain rule on *D* (*y*^{2} ) .)

3 (*x*^{2}+*y*^{2})^{2} ( 2*x* + 2 *y y*' ) = 8*x*^{2} (2 *y y*' ) + ( 16 *x* ) *y*^{2} ,

so that (Now solve for *y*' .)

6*x* (*x*^{2}+*y*^{2})^{2} + 6 *y* (*x*^{2}+*y*^{2})^{2}*y*' = 16 *x*^{2}*y y*' + 16 *x y*^{2} ,

6 *y* (*x*^{2}+*y*^{2})^{2}*y*' - 16 *x*^{2}*y y*' = 16 *x y*^{2} - 6*x* (*x*^{2}+*y*^{2})^{2} ,

(Factor out *y*' .)

*y*' [ 6 *y* (*x*^{2}+*y*^{2})^{2} - 16 *x*^{2}*y* ] = 16 *x y*^{2} - 6*x* (*x*^{2}+*y*^{2})^{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 *x*^{2} + (*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* ( *x*^{2} + (*y*-*x*)^{3} ) = *D* ( 9 ) ,

*D* ( *x*^{2} ) + *D* (*y*-*x*)^{3} = *D* ( 9 ) ,

2*x* + 3 (*y*-*x*)^{2}*D* (*y*-*x*) = 0 ,

2*x* + 3 (*y*-*x*)^{2} (*y*'-1) = 0 ,

so that (Now solve for *y*' .)

2*x* + 3 (*y*-*x*)^{2}*y*'- 3 (*y*-*x*)^{2} = 0 ,

3 (*y*-*x*)^{2}*y*' = 3 (*y*-*x*)^{2} - 2*x* ,

### 4.1 Implicit Differentiationap Calculus Algebra

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) .

### 4.1 Implicit Differentiationap Calculus Calculator

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*SOLUTION 12 :* Begin with *x*^{2}*y* + *y*^{4} = 4 + 2*x* . Now differentiate both sides of the original equation, getting

*D* ( *x*^{2}*y* + *y*^{4} ) = *D* ( 4 + 2*x* ) ,

*D* ( *x*^{2}*y* ) + *D* (*y*^{4} ) = *D* ( 4 ) + *D* ( 2*x* ) ,

( *x*^{2}*y*' + (2*x*) *y* ) + 4 *y*^{3}*y*' = 0 + 2 ,

so that (Now solve for *y*' .)

*x*^{2}*y*' + 4 *y*^{3}*y*' = 2 - 2*x y* ,

(Factor out *y*' .)

*y*' [ *x*^{2} + 4 *y*^{3} ] = 2 - 2*x 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*

### 4.1 Implicit Differentiationap Calculus Solver

*1998-06-23*