Question

What are the first and second derivatives of `f(x)=5^((x^5)-9x)`?

Answer 1

Following the rule to differentiate exponential functions: `(d(a^b))/(dx)=a^b(ln(a))b'`, we can proceed.

Explanation:

We'll rename `u=x^5-9x` so that `f(x)=5^u`

`(dy)/(dx)=5^u(ln(5))u'`

Substituting `u` and `u'`:

`(dy)/(dx)=5^(x^5-9x)ln(5)(5x^4-9)`

As for the second derivative, we must see we have a three terms product. To differentiate it, we'll consider two of them as one, in a sort of chain rule, and then derivate these two as well, as shown in the formula below:

`(abc)'=(ab)'c+(ab)c'=a'bc+ab'c+abc'`

Let's just identify things here. Following the general formula:
`color(red)(a=5^(x^5-9x))`, `color(blue)(b=ln(5))`, `color(green)(c=5x^4-9)`

`(dy^2)/(d^2x)=color(red)(5^(x^5-9x)ln(5)(5x^4-9))color(blue)((ln(5)))color(green)((5x^4-9))+cancel(color(red)(5^(x^5-9x))color(blue)((0))color(green)((5x^4-9)))+color(red)(5^(x^5-9x))color(blue)((ln(5)))color(green)((20x^3))`

As zero cancels the product of the second term, we have left:

`(dy^2)/(d^2x)=(5^(x^5-9x)ln(5))((5x^4-9)ln(5)+20x^3)`