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Find the derivative of $f(x)=-10\sqrt{x^{20}+9}$ with respect to $x$

I know to take the constant out and let $u=x^{20}+9$ \begin{align} f'=&-10 \cfrac{df}{dx}(u)^{1/2} \hspace{2cm} (1) \end{align} \begin{align} f'=&-10 \cfrac{1}{2}(u)^{-1/2} \hspace{2cm} (2) \\ f'=& \cfrac{-10}{2\sqrt{u}} \hspace{3.75cm} (3) \\ f' =& \cfrac{-5}{\sqrt{x^{20}+9}} \hspace{2.8cm} (4) \end{align}

I know this is very wrong, but I don't understand why the correct answer is $-10 \cdot \cfrac{1}{2\sqrt{x^{20}+9}}\cdot 20x^{19}$. I understand everything except the last term, $20x^{19}$. I'm aware it is the derivative of $u$, but I don't understand why we multiply by that to the numerator after having already taken the derivative of root $u$ as shown in line $2$.

can anyone explain why multiplying that term is necessary?

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    $\begingroup$ Did you use the chain rule properly? $\endgroup$ Dec 25 '19 at 5:21
  • $\begingroup$ $$\dfrac{d(f(x))}{dx} \ne \dfrac{df(u)}{dx}$$ $\endgroup$ Dec 25 '19 at 5:29
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$\frac d {dx} h(g(x))=h'(g(x)) g'(x)$ by Chain Rule. You forgot $g'(x)$. [Here $h(x)=-10\sqrt x$ and $g(x)=x^{20}+9$].

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There was an error in your application of the chain rule. This is the correct way of solving it.

Let, $$u = x^{20}+9$$

Then,

$$f'(x)=-10 \cfrac{d(u)^{\frac 12}}{dx}$$ $$f'(x)=-10 \cfrac{d(u)^{\frac 12}}{du} \cfrac{du}{dx}$$ $$f'(x)=-10 \cfrac{u^{\frac {-1}{2}}}{2} \cfrac{du}{dx}$$ And we know that $$\cfrac {du}{dx} = 20x^{19}$$ Hence, $$f'(x)= \cfrac {-5}{\sqrt {x^{20}+9}} 20x^{19}$$ Therefore, $$f'(x)= -10 \cfrac{1}{2 \sqrt {x^{20}+9}} 20x^{19}$$

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