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

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Answer 1 · 2017-08-22T08:51:17

$-3x^-4$ $sqrt(x-e^(2x)) +x^-3 1/2(x-e^(2x))^(-1/2)(1-2e^(2x))$

$f(x)g(x)=d/dxf(x)g(x)+f(x)d/dxg(x)$

So, we have $(1/x^3)=x^-3$

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

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

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

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

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

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

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