What is the derivative of ${ln e}^{2 x}$ $ln e^(2x)$ ?

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Answer 1 · 2016-10-03T00:33:40

$y = ln (u)$ $y=ln(u)$

$:. dy/(du)=1/u=1/(e^(2x))$

$u=e^(2x)$ , $:. (du)/(dx)=2e^(2x)$

Now, using the chain rule...

$(dy)/(du)*(du)/(dx)=1/(e^(2x))*2e^(2x)=2$

$:. (dy)/(dx)=2$

This all worked because:

When $y=e^(f(x))$

$(dy)/(dx)=f'(x)e^(f(x))$

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