How do you find the derivatives of #y=(x-1)^(x^2+1)# by logarithmic differentiation?
1 Answer
Mar 15, 2017
# dy/dx = (x-1)^(x^2+1){ (x^2+1)/(x-1) + 2xln (x-1) } #
Explanation:
We have:
# y = (x-1)^(x^2+1) #
If we take Natural Logarithms of both sides we get:
# ln y = ln { (x-1)^(x^2+1) } #
# " " = (x^2+1)ln (x-1) #
Now we can differentiate implicitly (along with the product rule and chain rule) to get:
# 1/ydy/dx = (x^2+1)(1/(x-1) * 1) + (2x)(ln (x-1)) #
# " " = (x^2+1)/(x-1) + 2xln (x-1) #
Giving:
# dy/dx = (x-1)^(x^2+1){ (x^2+1)/(x-1) + 2xln (x-1) } #
