Solutions to Implicit Differentiation Problems (2024)

SOLUTIONS TO IMPLICIT DIFFERENTIATION PROBLEMS

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 + 3y² y' = 0 ,

so that (Now solve for y' .)

3y² y' = - 3x² ,

and

Click HERE to return to the list of problems.

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

Click HERE to return to the list of problems.

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

Click HERE to return to the list of problems.

SOLUTION 4 : Begin with y = x² y³ + x³ y² . Differentiate both sides of the equation, getting

D(y) = D ( x² y³ + x³ y² ) ,

D(y) = D ( x² y³ ) + D ( x³ y² ) ,

(Use the product rule twice.)

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

y' = 3x² y² y' + 2x y³ + 2x³ y y' + 3x² y² ,

so that (Now solve for y' .)

y' - 3x² y² y' - 2x³ y y' = 2x y³ + 3x² y² ,

See Also

Study Guide - Implicit Differentiation

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

so that (Now solve for y' .)

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

(Factor out y' .)

1 - y - 3x² - 4x = (3y² + x + 2) y' ,

and

Click HERE to return to the list of problems.

SOLUTION 9 : Begin with . Clear the fractions by multiplying both sides of the equation by x³ y³ , getting

y⁴ + x⁴ = x⁵ y⁷ .

Now differentiate both sides of the equation, getting

D ( y⁴ + x⁴ ) = D ( x⁵ y⁷ ) ,

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

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

4 y³ y' + 4 x³ = x⁵ (7 y⁶ y' ) + ( 5 x⁴ ) y⁷ ,

so that (Now solve for y' .)

4 y³ y' - 7 x⁵ y⁶ y' = 5 x⁴ y⁷ - 4 x³ ,

(Factor out y' .)

y' [ 4 y³ - 7 x⁵ y⁶ ] = 5 x⁴ y⁷ - 4 x³ ,

and

Click HERE to return to the list of problems.

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

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

(Factor out y' .)

y' [ 6 y (x²+y²)² - 16 x² y ] = 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 ) )

y = x + 2 .

Click HERE to return to the list of problems.

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)² D (y-x) = 0 ,

2x + 3 (y-x)² (y'-1) = 0 ,

so that (Now solve for y' .)

2x + 3 (y-x)² y'- 3 (y-x)² = 0 ,

3 (y-x)² y' = 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 ) ) ,

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

Click HERE to return to the list of problems.

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

D ( x² y + y⁴ ) = D ( 4 + 2x ) ,

D ( x² y ) + D (y⁴ ) = D ( 4 ) + D ( 2x ) ,

( x² y' + (2x) y ) + 4 y³ y' = 0 + 2 ,

so that (Now solve for y' .)

x² y' + 4 y³ y' = 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) .

Click HERE to return to the list of problems.

About this document ...

Duane Kouba
1998-06-23

Solutions to Implicit Differentiation Problems (2024)

FAQs

How do I get better at implicit differentiation? ›

The trick to using implicit differentiation is remembering that every time you take a derivative of y, you must multiply by dy/dx. Furthermore, you'll often find this method is much easier than having to rearrange an equation into explicit form if it's even possible.

Find Out More ›

How to solve implicit functions? ›

First differentiate the entire expression f(x, y) = 0, with reference to one independent variable x. As a second step, find the dy/dx of the expression by algebraically moving the variables. The final answer of the differentiation of implicit function would have both variables.

Discover More ›

How do we apply implicit differentiation in real life situation? ›

Example 1: Engineering

In engineering, implicit differentiation can be used to determine how varying one physical property (such as pressure) might impact another (such as volume) when they are related through an equation that does not explicitly define one as a function of the other.

Get More Info Here ›

What is the formula for implicit differentiation? ›

The method of finding an implicit derivative is not so much a formula as it is a procedure, the steps for which will be detailed later in this article. However, a formula for implicit differentiation of y terms is, in essence, the Chain Rule: d u d x = d u d y ⋅ d y d x , where u is the term containing a y.

Learn More Now ›

What is the chain rule for implicit differentiation? ›

Implicit differentiation helps us find dy/dx even for relationships like that. This is done using the chain rule, and viewing y as an implicit function of x. For example, according to the chain rule, the derivative of y² would be 2y⋅(dy/dx).

Learn More Now ›

How to do differentiation step by step? ›

The differentiation rules are listed as follows:

Sum Rule: If y = u(x) ± v(x), then dy/dx = du/dx ± dv/dx.
Product Rule: If y = u(x) × v(x), then dy/dx = u.dv/dx + v.du/dx.
Quotient Rule: If y = u(x) ÷ v(x), then dy/dx = (v.du/dx- u.dv/dx)/ v²

More items...

What is the point of implicit differentiation? ›

Implicit differentiation is useful for, among other things, finding tangent and normal lines to a curve that cannot be expressed in the form y=f(x) y = f ( x ) .

View Details ›

How to differentiate with respect to y? ›

If x is a variable and y is another variable, then the rate of change of x with respect to y is given by dy/dx. This is the general expression of derivative of a function and is represented as f'(x) = dy/dx, where y = f(x) is any function.

Read The Full Story ›

How to solve a differential equation? ›

We can solve these differential equations using the technique of an integrating factor. We multiply both sides of the differential equation by the integrating factor I which is defined as I = e∫ P dx. ⇔ Iy = ∫ IQ dx since d dx (Iy) = I dy dx + IPy by the product rule.

Know More ›

What is an example of a differentiation of implicit function? ›

Chain Rule in Implicit Differentiation

We represent the derivative of two functions in product form. This can be understood from the following example: For example, we have to differentiate 3xy² implicitly with respect to x. d/dx(3xy) = 3y²+3x.

Read The Full Story ›

What is the application of implicit differentiation? ›

Implicit differentiation has an important application: it allows to compute the derivatives of inverse functions. It is good that we review this, because we can use these derivatives to find anti-derivatives.

Show Me More ›

What is an example of implicit? ›

Examples of implicit in a Sentence

There is a sense of moral duty implicit in her writings. I have implicit trust in her honesty. These examples are programmatically compiled from various online sources to illustrate current usage of the word 'implicit.

Know More ›

How do you set up implicit differentiation? ›

Here are the two basic implicit differentiation steps. Suppose you are differentiating with respect to x. Differentiate each side of the equation by treating y as an implicit function of x. This means you need to use the Chain Rule on terms that include y by multiplying by d y d x \frac{dy}{dx} dxdy.

Get More Info Here ›

What is the solution of a differential equation in implicit form? ›

A relation F(x,y)=0 is said to be an implicit solution of a differential equation involving x, y, and derivatives of y with respect to x if F(x,y)=0 defines one or more explicit solutions of the differential equation.

Learn More ›

What is the implicit difference equation? ›

In the case of deterministic, an implicit difference equation (IDE) can be described in the form AnX(n+1)=BnX(n)+qn,n∈N,(1) where A n , B n ∈ R d × d , X ( n ) , q n ∈ R d and A n may be a singular matrix.

Discover More ›