What is the derivative of $y = {(x - 2)}^{x + 1}$ $y=(x-2)^(x+1)$ ?

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Answer 1 · 2016-07-27T03:41:39

$(ln (x - 2) + \frac{x + 1}{x - 2}) {(x - 2)}^{x + 1}$ $( ln(x-2) + {x+1}/{x-2}) (x-2)^{x+1}$

The easiest way to find the derivative is to start with $ln y$ $ln y$ :

$ln y = (x + 1) ln (x - 2)$ $ln y = (x+1) ln(x-2)$

so that, differentiating :

$\frac{1}{y} \frac{d y}{d x} = ln (x - 2) + \frac{x + 1}{x - 2}$ $1/y {dy}/{dx} = ln(x-2) + {x+1}/{x-2}$

so that

$\frac{d y}{d x} = (ln (x - 2) + \frac{x + 1}{x - 2}) {(x - 2)}^{x + 1}$ ${dy}/{dx} = ( ln(x-2) + {x+1}/{x-2}) (x-2)^{x+1}$

What is the derivative of y=(x−2)x+1y=(x-2)^(x+1)?