Question

Consider the function.

f(x) = 1012x^101 – 72x^75 + pix^2 –e^2x + 100346

Compute f^102)(x). Give your answer in exact form.

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*2*1x^(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)) * g'(x)`

Then the first few derivatives of the exponential term are

`\left(e^(2x)\right)' = e^(2x) * (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)}`