# Derivative chainrule on khanacadamy ignoring some terms

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 "$g(x)$"?

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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... – Babak 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. – Babak S. Mar 1 '13 at 10:22
You could inform Khanacademy of this error so it won't confuse other students too. – Gibarian Mar 1 '13 at 12:56

$$f(x) = \sin(6x^2−4x−1+3x^{−1}−5x^{−2})$$
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 $\frac{dy}{dx} (x^n) = nx^{n-1}$.