Limitation first principle method

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Homework Statement


Should be a simple question but its really tough for me because my lecturer had never teach me before about x^3

Anyway, the question is find the first derivative by using first principle method :
c) x^3+2x^2+1


Homework Equations





The Attempt at a Solution


I did dy/dx = lim h->0 but during the process I am stuck because I am pretty confused about the expand part especially during the power for 3
 
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What do you mean by expanding for power 3? Assuming you want to expand (x+h)3, you need to multiply (x+h) thrice :wink: Or you can remember the result directly, or you can use the binomial theorem...

Edit : Do show your work, it'll be easier to help you out then :smile:
 
Last edited:
Yea, I meant by expanding the (x+h)3. I had managed to solve it using binomial theorem and it has saved me a lot of trouble. Thanks a lot Infinitum. Cheers :)
 
You're welcome! :smile:On a side note,
(a+b)3 is quite a commonly used identity, just like (a+b)2. It will surely help to memorize it.
 
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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