How do you differentiate $g (x) = (\frac{1}{x^{3}}) \cdot \sqrt{x - e^{2 x}}$ $g(x) = (1/x^3)*sqrt(x-e^(2x))$ using the product rule?

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How do you differentiate $g (x) = (\frac{1}{x^{3}}) \cdot \sqrt{x - e^{2 x}}$ $g(x) = (1/x^3)*sqrt(x-e^(2x))$ using the product rule?

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Answer 1 · 2017-08-22T08:51:17

$- 3 x^{- 4}$ $-3x^-4$ $\sqrt{x - e^{2 x}} + x^{- 3} \frac{1}{2} {(x - e^{2 x})}^{- \frac{1}{2}} (1 - 2 e^{2 x})$ $sqrt(x-e^(2x)) +x^-3 1/2(x-e^(2x))^(-1/2)(1-2e^(2x))$

Explanation:

$f (x) g (x) = \frac{d}{d x} f (x) g (x) + f (x) \frac{d}{d x} g (x)$ $f(x)g(x)=d/dxf(x)g(x)+f(x)d/dxg(x)$

So, we have $(\frac{1}{x^{3}}) = x^{- 3}$ $(1/x^3)=x^-3$

$\frac{d}{d x} x^{- 3} = - 3 x^{- 4}$ $d/dx x^-3=-3x^-4$ using this rule $n x^{n - 1}$ $nx^(n-1)$

$- 3 x^{- 4}$ $-3x^-4$ $\sqrt{x - e^{2 x}} + x^{- 3}$ $sqrt(x-e^(2x))+x^-3$

then the derivative of $\sqrt{x - e^{2 x}}$ $sqrt(x-e^(2x)$ $=$ $=$ ${(x - e^{2 x})}^{\frac{1}{2}}$ $(x-e^(2x))^(1/2)$

${(x - e^{2 x})}^{\frac{1}{2}}$ $(x-e^(2x))^(1/2)$ using chain rule enter image source here

${(x - e^{2 x})}^{\frac{1}{2}}$ $(x-e^(2x))^(1/2)$ = 1/2(x-e^(2x))^(-1/2)(1-2e^(2x))#

So, it's equal to
$- 3 x^{- 4}$ $-3x^-4$ $\sqrt{x - e^{2 x}} + x^{- 3} \frac{1}{2} {(x - e^{2 x})}^{- \frac{1}{2}} (1 - 2 e^{2 x})$ $sqrt(x-e^(2x)) +x^-3 1/2(x-e^(2x))^(-1/2)(1-2e^(2x))$

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How do you differentiate $g (x) = (\frac{1}{x^{3}}) \cdot \sqrt{x - e^{2 x}}$ $g(x) = (1/x^3)*sqrt(x-e^(2x))$ using the product rule?

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Explanation:

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How do you differentiate g(x)=(1x3)⋅√x−e2xg(x) = (1/x^3)*sqrt(x-e^(2x)) using the product rule?

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How do you differentiate $g (x) = (\frac{1}{x^{3}}) \cdot \sqrt{x - e^{2 x}}$ $g(x) = (1/x^3)*sqrt(x-e^(2x))$ using the product rule?