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

1 Answer
Jan 1, 2016

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)