Solving the Equation of cos^3(e^4(theta))

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Homework Statement



(cos^3)(e^4(theta))

Homework Equations


f(x)=f(g(x))


The Attempt at a Solution


u=e^49theta))

-3sin^2(e^4(theta)) x e^4(theta)


Just not sure what I'm doing wrong, I know what the answer is I'm just not sure how to get to it.
 
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For starters, you need to be clearer. Is your problem cos(e^4\theta)^3 ? It's hard to tell from the way you wrote it. If it is, then you need to use the chain rule, which is:

[f( g(x) )] ' = f'(g(x)) * g'(x).
 
The typical way to recall the steps if you're having trouble is:

1. derivative of the outside function, leaving the inside function untouched.
2. multiply by the derivative of the inside function.
 
the question is cos^3 * e^4theta the theta is connected to the raised power of 4.
 
You really need to use LaTeX or appropriately placed parentheses. My comments on the chain rule still apply. Where exactly are you getting stuck?
 
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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