# The composite function in the form f ( g ( x ) ) and obtain the derivative of y . ### 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 3.4, Problem 4E
To determine

## To find: The composite function in the form f(g(x)) and obtain the derivative of y.

Expert Solution

The inner function is u=cotx and the outer function is f(u)=sinu.

The derivative of y is dydx=csc2xcos(cotx)_.

### Explanation of Solution

Given:

The function is y=sin(cotx).

Formula used:

The Chain Rule:

If h is differentiable at x and g is differentiable at h(x), then the composite function F=gh defined by F(x)=g(h(x)) is differentiable at x and F is given by the product,

F(x)=g(h(x))h(x) (1)

Calculation:

Let the inner function be u=g(x) and the outer function be y=f(u).

Then, g(x)=cotx and f(u)=sinu. That is,

y=sin(cotx)=f(cotx)=f(g(x))

Therefore, y=f(g(x)).

Hence, the inner function is u=cotx and the outer function is f(u)=sinu.

Thus, the required form of composite function is f(g(x))=sinu.

Obtain the derivative of y is

Let h(x)=cotx and g(u)=sinu  where u=h(x).

Apply the chain rule as shown in equation (1),

y(x)=g(h(x))h(x) (2)

The derivative g(h(x)) is computed as follows,

g(h(x))=g(u)=ddu(g(u))=ddusinu=cosu

Substitute u=cotx in above equation,

g(h(x))=cos(cotx)

The derivative g(h(x)) is g(h(x))=cos(cotx).

The derivative of h(x) is computed as follows,

h(x)=ddx(cotx)=csc2x

Thus, the derivative of h(x) is h(x)=csc2x.

Substitute cos(cotx) for g(h(x)) and csc2x for h(x) in equation (2),

g(h(x))h(x)=[cos(cotx)][csc2x]=csc2xcos(cotx)

Therefore, the derivative of y=sin(cotx) is dydx=csc2xcos(cotx)_.

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers! 