Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I watched the chain rule series on khanacademy.org and decided to do the "questions". One of the questions is:

Let $y = \sin(6x^2−4x−1+3x^{−1}−5x^{−2})$

$dy/dx=?$

The answer is $dy/dx=(\cos(6x^2−4x−1+3x^{−1}−5x^{−2}))(12x−4)$

Since

$dy/dx [f(g(x))] = {\rm D}f(g(x))*{\rm D}g(x)$

I figure that it should be

$dy/dx = \cos(6x^2−4x−1+3x^{−1}−5x^{−2})(12x-4-3x^{-2}+10x^{-3})$

Why aren't they deriving the negative exponent terms in the inner function "$g(x)$"?

Thank you in advance for your help!

share|improve this question
    
wolfram alpha solves it as www.wolframalpha.com/input/?i=derivativ+sin(6x^2−4x−1%2B3x^−1−5x^−2) –  54N1 Mar 1 '13 at 9:27
    
If WA admitted what you had got, therefore they ignored some terms or... –  B. S. Mar 1 '13 at 9:58
    
Babak S. Why would they ignore terms? Are the terms somehow insignificant? –  54N1 Mar 1 '13 at 10:20
    
As it is written above I think they did. No! those terms are depend to $x$ and so they are important. –  B. S. Mar 1 '13 at 10:22
8  
You could inform Khanacademy of this error so it won't confuse other students too. –  Gibarian Mar 1 '13 at 12:56

1 Answer 1

It seems that there is some error in presenting the solution. That being said, the answer that you have provided is correct. The following is a step-by-step solution for future users.

$$f(x) = \sin(6x^2−4x−1+3x^{−1}−5x^{−2})$$

In order to find the derivative, we must use the chain rule. Therefore: $$f'(g(x)) * g'(x)$$ Where $f(x)$ is $\sin(x)$ and $g(x)$ is $6x^2−4x−1+3x^{−1}−5x^{−2}$. Therefore, the solution would be:

$$\begin{align} \cos (6x^2−4x−1+3x^{−1}−5x^{−2}) * 12x-4+3x^{-2}-5x^{-3} \end{align}$$

because the derivative of $\sin(x)$ is $\cos(x)$ and derivative of $g(x)$ can be found through the general forumla $\frac{dy}{dx} (x^n) = nx^{n-1}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.