Question

Consider the function.

Answer 1

It looks like you're given

$f(x) = 1012x^{101} - 72x^{75} + \pi x^2 - e^{2x} + 100346$

and are asked to find the 102nd derivative of f(x).

Recall the power rule: for integer n,

$\displaystyle \left(x^n\right)' = nx^{n-1}$

This means that the power of x reduces to 0 after differentiating n times, and you're left with a constant coefficient n! :

• after differentiating 2 times,

$\left(x^n\right)'' = \left(nx^{n-1}\right)' = n(n-1)x^{n-2}$

• after differentiating 3 times,

$\left(x^n\right)^{(3)} = \left(n(n-1)x^{n-2}\right)' = n(n-1)(n-2)x^{n-3}$

• and so on, up to the n-th time, which yields

$\left(x^n\right)^{(n)} = n(n-1)(n-2)\cdots\times2\times1x^{n-n} = n!$

As soon as you have a constant, the next derivative will be 0. This means that after differentiating 102 times, the first 3 terms of f(x), as well as the constant term, will vanish.

Recall the chain rule:

$\bigg(f(g(x))\bigg)' = f'(g(x)) \times g'(x)$

Then the first few derivatives of the exponential term are

$\left(e^{2x}\right)' = e^{2x} \times (2x)' = 2e^{2x}$

$\left(e^{2x}\right)'' = 2\left(e^{2x}\right)' = 2^2e^{2x}$

$\left(e^{2x}\right)^{(3)} = 2^2\left(e^{2x}\right)' = 2^3e^{2x}$

and so on, with n-th derivative

$\left(e^{2x}\right)^{(n)} = 2^ne^{2x}$

Putting everything together, we have

$\boxed{f^{(102)}(x) = -2^{102}e^{2x}}$