Can I Use the Power Rule to Get the Derivative Here?

Danatron
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can i use the power rule to get the derivative here?

f ' (x) = 3x^2 - 2(2x^1) + 1
 
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Of course, why not? Do you use any other rule for that?
Your answer is correct
 
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ok good, so i wouldn't go again until there were no powers?

eg. f ' (x) = 2(3x) - 2(2x) + 1
 
Danatron said:
ok good, so i wouldn't go again until there were no powers?

eg. f ' (x) = 2(3x) - 2(2x) + 1
No. Why do you have to differentiate it again? The notation ##f'(x)## means-Differentiate once. Similarly, the notation ##f''(x)## means differentiate twice.

In leibniz notations ##\frac{\text{d}}{\text{d}x}## means- Differentiate once and ##\frac{\text{d}^2}{\text{d}x^2}## means- differentiate twice and so on.

In your example,(differentiating twice) you should write ##f''(x)## and this should be equal to the derivative of ##3x^2 - 4x + 1## which is ##6x+4##
Note that the derivative of a constant (1) is zero.

So in general, we differentiate the function ##f##, and again differentiate the derivative of the function ##f##, to differentiate ##f## twice. This can also be extended to differentiating 10000 times :wink:
 
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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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