Help finding the higher order derivative!!?

Question

Given f'(x)=x^2 find f''(x)


please help

Answer 1

f'(x) = LIM (f(x + delta x) - f(x))/ delta x
as delta x approaches 0

f(x + delta x) = (x + delta x)^2 = x^2 + 2*x*delta x + (delta x)^2
f(x + delta x) - f(x) = 2*x*delta x + (delta x)^2 (we subtracted x^2)
now divide by delta x
2x + delta x
Since delta x = 0 at the limit this leaves us with
f'(x) = 2x

Answer 2

To find higher order derivative of any function differentiate the derivative of the function.

Thus to find f"(x) where f(x) = x^2,

We first find f'(x) = 2x (From 2x^(2-1) )
therefore, to find f" , we simply diferentiate 2x (the derivative of our given function) again:

Thus f" = 2 (from 2*1x^(1-1)).

That's it but if you want to go further, you can find f"' which will be the derivative of 2, that is f"' = 0 (from derivative of a constatnt).

Hope this helps.

Answer 3

f''(x) = f'(f'(x))
f''(x) = 2x aka d2y/dx2 (2s superscripted)

Just keep differentiating the previous derivative to find the next one.

Answer 4

f(x) = x^2

First derivative f'(x) = 2x

Second derivative f''(x) = 2

All subequent derivatives are equal to zero

Answer 5

f''(x) = d/dx(f'(x)) = d/dx(x^2) = 2x

Answer 6

f'(x)=x^2
f"(x)=2x