How do I find #f'(x)# for #f(x)=5^x# ?
4 Answers
Apr 4, 2018
#dy/dx = 5^x log5 #
Explanation:
#y=5^x#
#logy=x log5#
#dy/dx. 1/y=x.(0)+log5.(1)#
#dy/dx . 1/y =log5#
#dy/dx =log5 . y#
#dy/dx =log5 . 5^x#
#dy/dx = 5^x log5 #
Apr 4, 2018
Taking log on both the sides,
Now, applying implicit differentiation,
Apr 4, 2018
Explanation:
Let
then
or
Apr 4, 2018
Explanation:
Given:
Using the common integral that