Differentiate each of the following from first principles:
x3 + 4x2 + 3x + 2
We need to find the derivative of f(x) = x3 + 4x2 + 3x + 2
Derivative of a function f(x) from first principle is given by –
f’(x) = {where h is a very small positive number}
∴ derivative of f(x) = x3 + 4x2 + 3x + 2is given as –
f’(x) =
⇒ f’(x) =
⇒ f’(x) =
Using (a + b)2 = a2 + 2ab + b2 , we have –
⇒ f’(x) =
⇒ f’(x) =
⇒ f’(x) =
Take h common –
⇒ f’(x) =
As h is cancelled, so there is no more indeterminate form possible if we put value of h = 0.
So, evaluate the limit by putting h = 0
⇒ f’(x) =
⇒ f’(x) = 3x2 + 3x(0) + 8x + 3 + 02 + 4(0)
⇒ f’(x) = 3x2 + 8x + 3
Hence,
Derivative of f(x) = x3 + 4x2 + 3x + 2 is 3x2 + 8x + 3