How do you differentiate ${(e^{2 x} + 2 x)}^{0.5}$ $(e^(2x) + 2x) ^0.5$ ?

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How do you differentiate ${(e^{2 x} + 2 x)}^{0.5}$ $(e^(2x) + 2x) ^0.5$ ?

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Answer 1 · 2016-01-13T21:26:24

$\frac{e^{2 x} + 1}{{(e^{2 x} + 2 x)}^{\frac{1}{2}}}$ $(e^(2x) + 1)/(e^(2x) + 2x )^(1/2)$

Explanation:

$f (x) = {(e^{2 x} + 2 x)}^{\frac{1}{2}}$ $f(x) = (e^(2x )+ 2x )^(1/2)$

using the chain rule :(twice)

$f'(x) = 1/2(e^(2x) + 2x )^(-1/2) . d/dx (e^(2x) + 2x )$

$= 1/2(e^(2x) + 2x )^(-1/2) (e^(2x) .d/dx(2x) + 2 )$

$= 1/2(e^(2x) +2x )^(-1/2) (2e^(2x) + 2 )$

$= 1/2 (e^(2x) +2x )^(-1/2) .2 ( e^(2x) + 1 )$

$rArrf'(x) = (e^(2x) + 1 )/(e^(2x) + 2x )^(1/2)$

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How do you differentiate ${(e^{2 x} + 2 x)}^{0.5}$ $(e^(2x) + 2x) ^0.5$ ?

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Explanation:

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How do you differentiate (e^(2x) + 2x) ^0.5(e2x+2x)0.5 (e^(2x) + 2x) ^0.5?

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Explanation:

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How do you differentiate ${(e^{2 x} + 2 x)}^{0.5}$ $(e^(2x) + 2x) ^0.5$ ?