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:

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

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