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.

• yes it is @Amzoti – user157908 Jun 20 '14 at 4:36
• @Amzoti whats the u and the d :? – user157908 Jun 20 '14 at 4:45

$$p'(t)=3\cdot\frac{d}{dt}(e^{-2e^{2t}}).$$ Let $u=-2e^{2t}$, therefore $$\frac{du}{dt}=-2(2\cdot e^{2t})=-4e^{2t}.$$ Therefore we have $$p'(t)=3\cdot \frac{d}{du}e^u\cdot-4e^{2t}$$ $$=3e^u \cdot -4e^{2t}$$ $$=-12e^{-2e^{2t}}\cdot e^{2t}$$ $$=-12e^{2t-2e^{2t}}.$$

The derivative is

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

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

Hence,

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

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.

it's basically just the chain rule.

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