How do you differentiate this equation?

#y=(a^x-b^x)/(a^x-c^x)#

1 Answer
Ananda Dasgupta
Apr 13, 2018

# (a^x(b^x-c^x) ln a-b^x(a^x-c^x) ln b+(a^x-b^x)c^xln c)/(a^x-c^x)^2 #

Explanation:

The derivative of #a^x# is #a^x ln a#.

We can use the quotient rule to calculate the required derivative.

# d/dx( (a^x-b^x)/(a^x-c^x)) #
#= ((a^x-c^x)d/dx(a^x-b^x)-(a^x-b^x)d/dx(a^x-c^x))/(a^x-c^x)^2#
#= ((a^x-c^x)(a^x ln a-b^x ln b)-(a^x-b^x)(a^xln a-c^xln c))/(a^x-c^x)^2#
#= (a^x(b^x-c^x) ln a-b^x(a^x-c^x) ln b+(a^x-b^x)c^xln c)/(a^x-c^x)^2 #

Related questions
Impact of this question
472 views around the world
You can reuse this answer
Creative Commons License