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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?

Dylan
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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.

Caedmon
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  • so it could also be: $$-\frac{21\sqrt{2}}{4x\sqrt{x^9}}$$ ? – Dylan Jan 13 '22 at 20:37
  • @DylanWeijgertze Oh no, that's not the same. $\sqrt{x^7} = x^\frac{7}{2}$ and when you multiply the same variable, the exponents add: $x^3 * x^2 = x^5$, and $x^1 * x^\frac{7}{2} = x^\frac{9}{2} = \sqrt{x^9}$. So $x \sqrt{x^9} = \sqrt{x^{11}}$. – Caedmon Jan 13 '22 at 21:46
  • @DylanWeijgertze Does that make more sense? – Caedmon Jan 13 '22 at 21:48
  • Oh so it would be $$-\frac{21\sqrt{2}}{4\sqrt{x^9}}$$ instead of

    $$-\frac{21\sqrt{2}}{4x\sqrt{x^7}}$$

     – Dylan Jan 14 '22 at 09:26
  • @DylanWeijgertze Yes those two equations are identical. If you wanna check plug them both into a graphing calculator and you'll see there is no difference between them. – Caedmon Jan 14 '22 at 13:51