<s>[INST] Get derivative with square root in the denominator Given:
$$\begin{align}
f(x) &= \frac{3}{\sqrt{2x^7}} \\
\\
\frac{df}{dx} &=
\end{align}$$
The expected answer is:
$$\begin{align}
f(x) &= \frac{3}{\sqrt{2x^7}} \\
\\
\frac{df}{dx} &= -\frac{21\sqrt{2}}{4x\sqrt{x^7}}
\end{align}$$
But what steps would I need to take to that answer?
 [/INST] This would use the chain rule. Rewrite the function as:
$$f(x) = 3(2x^7)^{-\frac{1}{2}}$$
Using the chain rule:
$$\frac{df}{dx} = -\frac{1}{2} *3(2x^7)^{-\frac{3}{2}} * 14x^6$$
$$\frac{df}{dx} = -\frac{21x^6}{2^\frac{3}{2} x^\frac{21}{2}}$$
$$\frac{df}{dx} = -\frac{21\sqrt{2}}{\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{2}x^\frac{9}{2}}$$
$$\frac{df}{dx} = -\frac{21\sqrt{2}}{4x^\frac{9}{2}}$$
$$\frac{df}{dx} = -\frac{21\sqrt{2}}{4x \sqrt{x^7}}$$
This is really really weird simplification, like why write $x^\frac{9}{2}$ as $x\sqrt{x^7}$ instead of $\sqrt{x^9}$? Idk but that's how it was done.
</s>