# The derivative d x d y by the use of implicit differentiation.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.5, Problem 20E
To determine

## To find: The derivative dxdy by the use of implicit differentiation.

Expert Solution

The derivative of dxdy by the use of implicit differentiation is dxdy=xsec2ysecxysecxtanxtany.

### Explanation of Solution

Given:

The equation is ysecx=xtany.

Derivative rules:

(1) Chain rule: If y=f(u) and u=g(x) are both differentiable function, then

dydx=dydududx.

(2) Product Rule: If f1(x) and f2(x) are both differentiable, then

ddx(f(x)g(x))=f(x)ddx(g(x))+g(x)ddx(f(x)).

Calculation:

Let y be an independent variable and x be a dependent variable.

The given equation is,

ysecx=xtany

Differentiate implicitly with respect to y on both sides of the above equation,

ddy(ysecx)=ddy(xtany)

Apply the product rule (2) and simplify the terms,

yddy(secx)+secxddy(y)=xddy(tany)+tanyddy(x)yddy(secx)+secx(1)=x(sec2y)+tanydxdy

Apply the chain rule (1) as,

y(ddx(secx)dxdy)+secx(1)=x(sec2y)+tanydxdyysecxtanxdxdy+secx=x(sec2y)+tanydxdy

Combine the terms containing dxdy and obtain the expression for dxdy.

ysecxtanxdxdytanydxdy=x(sec2y)secxdxdy(ysecxtanxtany)=xsec2ysecxdxdy=xsec2ysecxysecxtanxtany

Therefore, the derivative of dxdy by the use of implicit differentiation is dxdy=xsec2ysecxysecxtanxtany.

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