Question

How do you differentiate `y=(x-2)^(x+1)`?

Answer 1

`dy/dx= (x - 2)^(x + 1)((x- 2)ln(x - 2) + x + 1)/(x- 2)`

Explanation:

Take the natural logarithm of both sides.

`lny = ln(x- 2)^(x + 1)`

Now apply `lna^n = nlna`.

`lny= (x + 1)ln(x - 2)`

Distribute:

`lny = xln(x - 2) + ln(x- 2)`

Now differentiate using the product rule and implicit differentiation. Let `u = x` and `v = ln(x - 2)`. By the product rule `d/dx(uv) = u'v + v'u`. The derivative of `v`, by the chain rule, is `1/(x - 2)` and `u` is `1`.

`1/y(dy/dx) = 1ln(x - 2) + x/(x - 2) + 1/(x - 2)`

`1/y(dy/dx) = ((x - 2)ln(x - 2) + x + 1)/(x - 2)`

`dy/dx = y((x - 2)ln(x - 2) + x + 1)/(x- 2)`

`dy/dx= (x - 2)^(x + 1)((x- 2)ln(x - 2) + x + 1)/(x- 2)`

Hopefully this helps!