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How to solve this derivative?

$$\large p(t) = 3e^{{-2e}^{2t}}$$

It looks weird to have two exponents instead of one. I tried to solve it but i got stuck.

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  • $\begingroup$ yes it is @Amzoti $\endgroup$ – user157908 Jun 20 '14 at 4:36
  • $\begingroup$ @Amzoti whats the u and the d :? $\endgroup$ – user157908 Jun 20 '14 at 4:45
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$$p'(t)=3\cdot\frac{d}{dt}(e^{-2e^{2t}}).$$ Let $u=-2e^{2t}$, therefore $$\frac{du}{dt}=-2(2\cdot e^{2t})=-4e^{2t}.$$ Therefore we have $$p'(t)=3\cdot \frac{d}{du}e^u\cdot-4e^{2t}$$ $$=3e^u \cdot -4e^{2t}$$ $$=-12e^{-2e^{2t}}\cdot e^{2t}$$ $$=-12e^{2t-2e^{2t}}.$$

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The derivative is

$$p'(t) = 3e^{f(t)}f'(t)$$

where $f(t) = -2e^{2t}$ and $f'(t)=-4e^{2t}$.

Hence,

$$p'(t) = -12e^{-2e^{2t}}e^{2t}$$

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You look for the derivative of $$p = 3e^{{-2e}^{2t}}$$ Logarithmic differentiation is a nice way to do it. $$\log (p)=\log (3)-2e^{2t}$$ then $$\frac {p'}{p}=-4e^{2t}$$ and now muliply the rhs by $p$ to get the result given in previous answers.

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it's basically just the chain rule.

$$\frac{d}{dt}3e^{\displaystyle-2e^{\displaystyle2t}} = \frac{d}{d (-2e^{2t})}3e^{\displaystyle-2e^{\displaystyle2t}}\cdot\frac{d}{d(2t)}(-2e^{2t})\cdot\frac{d}{t}2t$$ $$= (3e^{\displaystyle-2e^{\displaystyle2t}})( -2e^{2t})(2)$$ $$= -12e^{-2e^{2t}+2t}$$

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