What is the derivative of f(x) = 2cos3x + 3sin2x on the interval (-pie, pie)?

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Homework Statement



find the maxima and minima of the function f(x)=2cos3x + 3sin2x on the interval (-pie, pie)

Homework Equations





The Attempt at a Solution

 
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You must show your work, before we can help you. What have you tried> Maybe something involving the derivative?
 
im not sure if i did the derivative right. i got -2sin3x+ 3cosx as the first derivative
 
ryan.1015 said:
im not sure if i did the derivative right. i got -2sin3x+ 3cosx as the first derivative

I'm sure that you didn't. You need to use the chain rule. Your answer should start with "f'(x) = "
 
pere callahan and mark44, is the following statement right..
find derivative of f(x) as below
f'(x) = -6sin(3x) + 6cos(2x)
now solve it by equating to zero. then find critical points. from these points and end points (pie and -pie) find the absolute maxima and minima & relative maxima and minima
 
ElectroPhysics said:
pere callahan and mark44, is the following statement right..
find derivative of f(x) as below
f'(x) = -6sin(3x) + 6cos(2x)
It is.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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