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

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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.
 
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parents said:

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) \ ?
 
SammyS said:
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
 
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'
 
Last edited:
Your answer looks good to me.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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