What rate is x^4 increasing when x=4?

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If x is positive and increasing at a rate of 2 units per second, at what rate is x^4 increasing when x=4?

I don't even know how to start.
If you can help me I would be greatful.
 
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If x is increasing at the rate of 2 units/second then it must be increasing wrt some variable, call it t. So you are given dx(t)/dt=2 and x(t)=4. Now take the derivative of x(t)^4 and don't forget the chain rule.
 
duh stupid me I know what to do thank you anyway
 

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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