How do I find the derivative of cos(x)^(x+7)?

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Help! I haven't the slightest clue on how to do this...

thanks in advance!
 
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MrGoodyear812 said:
Help! I haven't the slightest clue on how to do this...

thanks in advance!

y=cos^{x+7}(x)

Start by taking the natural logarithmic of both sides.
 
even the y?

so i'd have:

ln(y) = (x+7)ln(cos(x))

how do i get rid of the ln on the y?

cause i knew it was a ln problem, just i don't know how to get rid of the ln on the y
 
Don't worry about 'getting rid of the ln' yet. The problem is about taking derivatives, so take one and see what happens.
 
If you take the derivative of the left side, you should get (1/y)(dy/dx). On the right side, you should have the derivative of ((x + 7)ln(cos x)). Solve the resulting equation for dy/dx.
 

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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