How would I differentiate functions with square roots such as:
$$3\sqrt{x}-\frac{2}{x^2}$$
$$\sqrt{1-x^2}$$
$$\sqrt{\frac{1-2x}{1+2x}}$$
$$\frac{2x}{\sqrt{x^2+7}}$$
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4Hint: $\displaystyle \sqrt x=x^\frac{1}{2}$ – Sujit Bhattacharyya Jun 13 '19 at 04:14
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1Welcome to Math Stack Exchange. You could use the power rule, along with product, quotient, and/or chain rule. Cf. this question – J. W. Tanner Jun 13 '19 at 04:16
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I understand that x^1/2 is the same as √x and I've managed to work out the first two, but I'm still stuck on the last two. due too the division. I just don't know where ot start – jakeymaths Jun 13 '19 at 06:13
3 Answers
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$\dfrac{\operatorname d}{\operatorname dx}\sqrt x=\dfrac 12x^{-\frac12}$.
The rest is just chain rule etc.
For example: $(\sqrt{\dfrac{1-2x}{1+2x}})'=\dfrac 12(\dfrac {1-2x}{1+2x})^{-\frac12}\cdot(\dfrac{1-2x}{1+2x})'=\dfrac 12(\dfrac {1-2x}{1+2x})^{-\frac12}\cdot(\dfrac{-2(1+2x)-(1-2x)\cdot2}{(1+2x)^2})=\sqrt{\dfrac {1+2x}{1-2x}}\cdot(\dfrac{-2}{(1+2x)^2})$.
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You can take the general case, $$\frac{\mathrm{d}}{\mathrm{d}x}f(x)^a=af(x)^{a-1}\frac{\mathrm{d}}{\mathrm{d}x}f(x)$$
$$\sqrt{f(x)}=f(x)^{\frac{1}{2}}$$
$$\frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$$
Hence, $$\left(\sqrt{\frac{1-2x}{1+2x}}\right)' = \left[\left(\frac{1-2x}{1+2x}\right)^{\frac{1}{2}}\right]'=\frac{1}{2}\left(\frac{1-2x}{1+2x}\right)^{\frac{1}{2}-1}\left(\frac{1-2x}{1+2x}\right)'$$
Alexandru
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When you face products, quotients, powers, logarithmic differentiation makes life easier.
Consider the general case of $$g(x)=[f(x)]^a\implies \log(g(x))=a \log(f(x))$$ Differentiate both sides $$\frac{g'(x)}{g(x)}=a\frac{f'(x)}{f(x)}\implies g'(x)=a\ g(x)\frac{f'(x)}{f(x)}=a\ f(x)^{a-1} f'(x)$$
Claude Leibovici
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