Derivative of sin(x)/(1+x^2) using Chain Rule | Simple Homework Example

[parents]

Homework Statement

Use chain rule to find the derivative of f(x)= sin(x)/(1+x^2)

Homework Equations

Chain Rule (f(g(x)))'*g'(x)

The Attempt at a Solution

y'(x)= cos (x)/(1+x^2)* (1-x^2)/((1+x^2)^2)

I just want to make sure I am doing it correctly and this would be acceptable as a final answer.

 

[SammyS]

Homework Statement

Use chain rule to find the derivative of f(x)= sin(x)/(1+x^2)

Homework Equations

Chain Rule (f(g(x)))'*g'(x)

The Attempt at a Solution

y'(x)= cos (x)/(1+x^2)* (1-x^2)/((1+x^2)^2)

I just want to make sure I am doing it correctly and this would be acceptable as a final answer.

Hello parents. Welcome to PF !

Is the function $\ \displaystyle f(x)=\frac{\sin(x)}{1+x^2} \,,$

or is it $\ \displaystyle f(x)=\sin\left(\frac{x}{1+x^2}\right) \ ?$

 

[parents]

Hello parents. Welcome to PF !

Is the function $\ \displaystyle f(x)=\frac{\sin(x)}{1+x^2} \,,$

or is it $\ \displaystyle f(x)=\sin\left(\frac{x}{1+x^2}\right) \ ?$

Sorry! I see how that can be confusing. It's $\ \displaystyle f(x)=\sin\left(\frac{x}{1+x^2}\right)$

I am working on trying to put in equations correctly

 

[iRaid]

Make the equation f(u)=sin(u). Then take the derivative of sin(u) then multiply by the derivative of u.

So:
f'(u)=sin(u)'u'

 

[MostlyHarmless]

Your answer looks good to me.