How do you find the derivatives of #y=(x-1)^(x^2+1)# by logarithmic differentiation?

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Answer 1 · 2017-03-15T21:55:47

# dy/dx = (x-1)^(x^2+1){ (x^2+1)/(x-1) + 2xln (x-1) } #

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) } #