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I'm having trouble working out how to differentiate expressions that have both division and square roots in them such as:

$$\frac{2x}{\sqrt{x^2+7}}$$

if someone could show me the necessary process that would be great.

Jacob Dodd
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    Use $$\bigg(\frac{u}{v}\bigg)' = \frac{vu' - uv'}{v^2}$$ – 19aksh Jun 14 '19 at 03:16
  • sorry but could you explain that formula for me? – Jacob Dodd Jun 14 '19 at 03:17
  • The $'$ represents the derivative. Here consider $u = 2x$ and $v = \sqrt{x^2+7}$, find their derivatives and substitute in the above formula. Also apply chain rule for $v'$, so that $v' = \frac{1}{2\sqrt{x^2+7}}\cdot(2x)$ – 19aksh Jun 14 '19 at 03:22
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    It's the quotient rule for derivatives – J. W. Tanner Jun 14 '19 at 03:31
  • I got this : $$\frac{\sqrt{x^2+7}2-2x\frac{1}{2\sqrt{x^2+7}}}{\sqrt{x^2+7}}$$. I'm not sure if this is even correct but what do I do from here to reach the final derivative? – Jacob Dodd Jun 14 '19 at 03:38
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    There are 2 mistakes. $u= 2x \implies u' = 2$(correct) and $$v = \sqrt{x^2+7} \implies v' = \frac{2x}{2\sqrt{x^2+7}} = \frac{x}{\sqrt{x^2+7}}$$. Also, $v^2 = x^2+7$. So, $$(u/v)' = \frac{\sqrt{x^2+7}\cdot2 - 2x\frac{\color{red}{x}}{\sqrt{x^2+7}}}{\color{red}{x^2+7}} = \frac{2(x^2+7)-2x^2}{(x^2+7)\sqrt{x^2+7}}$$ Can you complete now? – 19aksh Jun 14 '19 at 03:49
  • Do I just need to further simplify? – Jacob Dodd Jun 14 '19 at 04:44
  • This is virtually the same as these two questions that were asked (and closed) yesterday: https://math.stackexchange.com/questions/3260661/differentiating-expressions-with-division-in-them, https://math.stackexchange.com/questions/3260587/how-to-differentiate-functions-with-square-roots – Hans Lundmark Jun 14 '19 at 07:35

2 Answers2

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Using the quotent rule $\bigg(\frac{u}{v}\bigg)' = \frac{vu' - v'u}{v^2}$

$$\frac{d}{dx}\left(\frac{2x}{\sqrt{x^2+7}}\right) = \frac{\sqrt{x^2+7} \cdot (2x)^{'}-2x\cdot {(\sqrt{x^2+7})^{'}}}{(\sqrt{x^2+7})^2}$$

$$=\frac{\sqrt{x^2+7} \cdot 2 - 2x\cdot \frac{1}{2\sqrt{x^2+7}}\cdot(2x) }{x^2+7}=\frac{2\sqrt{x^2+7}-\frac{2x^2}{\sqrt{x^2+7}}}{x^2+7}=\frac{\frac{2(x^2+7)-2x^2}{\sqrt{x^2+7}}}{x^2+7}=\boxed{\frac{14}{(x^2+7)\sqrt{x^2+7}}}$$

Vineet
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For this kind of problems where you have products, quotients and powers, logarithmic differentiation makes life easier.

Making the problem more general, consider $$f(x)=\frac{[g(x)]^a}{[h(x)]^b}\implies \log(f(x))=a \log(g(x))-b \log(h(x))$$ Differnetiate both sides to get $$\frac{f'(x)}{f(x)}=a\frac{g'(x)}{g(x)}-b\frac{h'(x)}{h(x)}\implies f'(x)=f(x)\Big(a\frac{g'(x)}{g(x)}-b\frac{h'(x)}{h(x)} \Big)$$ and simplify to finally obtain $$f'(x)=a g(x)^{a-1} h(x)^{-b} g'(x)-b g(x)^a h(x)^{-b-1} h'(x)$$

Claude Leibovici
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