Derivative of 1-e^2x: Simplified Solution

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derivative of 1-e^2x?

derivative of 1-e^2x?

i know that the answer is -2e^2x but i have no idea how to get there
 
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What does the "Chain Rule" say about taking derivatives?

hint: it says \frac{d}{dx}[e^u]=e^u*\frac{du}{dx}
 
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im not even sure if this requires the chain rule... could someone just show me what i need to do because I am really stuck.
 
erjkism said:
im not even sure if this requires the chain rule... could someone just show me what i need to do because I am really stuck.

The chain rule does just show you what to do. And so did I.

Casey
 
thanks i got it now.
 
What if it were just 1-e^u What is the derivative?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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