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)#