What is the rate of change of power?

DespicableMe
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Homework Statement



POwer P in watts, produced by an engine is P(R) = 130R / [(R+0.8)2 ] where R is resistance. What is the rate of change of power when R = 5?

Homework Equations



The Attempt at a Solution



I do not understand how the rate of change for this is

130/ ( (R+0.8)2) - 2(130R) / ((R+0.8)3)

I have tried plugging in R=5 into what I just typed above and I got the right answer. I just don't understand how they got to that equation.

One site wrote that the ROC of current with respect to resistance is 125/R2, for example, if current was 125/R

Can someone please clarify?
 
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Well, what you typed there is the derivative of P(R) with respect to R. I think that's what they are asking for. What are you confused about?
 
Dick said:
Well, what you typed there is the derivative of P(R) with respect to R. I think that's what they are asking for. What are you confused about?


They simplified the form of the quotient rule, so I wasn't sure why the whole thing was over the denominator to the power of 3 instead of the power of 4.

When I did the quotient rule again, I understood where I went wrong and how they simplified it, so I have the answer now. Thanks.
 

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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