You can easily derive it from formula for the derivative of a power, if you remember that the cubic root of x is equal to x1/3.
This question asks for the proof of the derivative, not the derivative itself.
Using the definition of derivative, lim f(x) as h approaches 0 where f(x) = (f(a+h)-f(a))/h, we get the following:
[(a+h)1/3 - a1/3]/h
Complete the cube with (a2 + ab + b2)
Multiply by [(a+h)2/3 + (a+h)1/3 × a1/3 + a2/3] / [(a+h)2/3 + (a+h)1/3 × a1/3 + a2/3]
This completes the cube in the numerator, resulting in the following:
(a + h - a) / (h × [(a+h)2/3 + (a+h)1/3 × a1/3 + a2/3])
h / (h × [(a+h)2/3 + (a+h)1/3 × a1/3 + a2/3])
h cancels
1 / [(a+h)2/3 + (a+h)1/3 × a1/3 + a2/3]
Now that we have a function that is continuous for all h, we can evaluate the limit by plugging in 0 for h.
This gives
1/[a2/3 + a1/3 × a1/3 + a2/3]
Simplify a1/3 × a1/3
1/[a2/3 + a2/3 + a2/3]
(1/3)a2/3 or (1/3)a-2/3
This agrees with the Power Rule.
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f'(x)=.5x^-.5 f''(x)=-.25x^-1.5 f'''(x)=.375x^-2.5 or f'''(x)=(3/8)x^(-5/2)
Derivative of lnx= (1/x)*(derivative of x) example: Find derivative of ln2x d(ln2x)/dx = (1/2x)*d(2x)/dx = (1/2x)*2===>1/x When the problem is like ln2x^2 or ln-square root of x...., the answer won't come out in form of 1/x.
0
It is x to the sixth power.
Derivative with respect to 'x' of (5x)1/2 = (1/2) (5x)-1/2 (5) = 2.5/sqrt(5x)