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I am following this tutorial, and I couldn't understand 2 mathematical rules, I am looking for a very simple, step-by-step logic please:

The Power Rule: $\frac{d}{dx} u^n = nu^{n-1}\frac{du}{dx}$

The Chain Rule: $\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$

P.S And yes, I did search but could not find fair explanation.

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  • $\begingroup$ Should be small f in chain rule. Also, u is a function of x, it is called g(x) in the chain rule. Note $f(g(x))=g(x)^n$ in that case. $\endgroup$ – Emil Jul 11 '18 at 12:18
  • $\begingroup$ In your statement of the chain rule, the capital $F$ should be a lower case $f$. What is it about the chain rule that you don't understand? How to derive it? How to use it? Can you use the chain rule to compute the derivative of the function $h(x) = \sin(x^2)$? $\endgroup$ – littleO Jul 11 '18 at 12:18
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    $\begingroup$ @Emil I fixed the typo. $\endgroup$ – Yahya Jul 11 '18 at 12:20
  • $\begingroup$ @littleO Let's focus on the Power Rule please, as far as I know: $x^n = nx^{n-1}$ But how come we multiply by du again ! $\endgroup$ – Yahya Jul 11 '18 at 12:21
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    $\begingroup$ The statement of the first rule might be more clear with different notation. Suppose that $h(x) = u(x)^n$. Then $h'(x) = n u(x)^{n-1} u'(x)$. This result can be shown using the chain rule, as follows. Notice that $h(x) = f(u(x))$, where $f(y) = y^n$. The chain rule tells us that $h'(x) = f'(u(x)) u'(x) = n u(x)^{n-1} u'(x)$. $\endgroup$ – littleO Jul 11 '18 at 12:27
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If $u=u(x)$ then \begin{align} \frac{d}{dx}u^n &= \frac{d}{du}u^n\frac{du}{dx}\\ &= n u^{n-1}\frac{du}{dx} \end{align}

Also, let $y=g(x)$ then \begin{align} \frac{d}{dx}f(y) &= \frac{dy}{dx}\frac{d}{dy}f(y) \\ &= g'(x)f'(y)\\ &=g'(x)f'(g(x)) \end{align}

Reference

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I'll use Taylor expansion to define the derivative: $f'$ is the only number (/linear application) such that: $f(x+h)-f(x)-f'(x)h=\omicron(h)$

One have: $f(g(x+h))=f(g(x)+g'(x)h+\omicron(h))=f(g(x))+f'(g(x))\bigl(g'(x)h+\omicron(h)\bigr) +\omicron(h)$

Hence the desired result: $f(g(x+h))-f(g(x))-f'(g(x))g'(x)h=\omicron(h)$

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