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

Answer 1

Aurora Alvarado

#( 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 = (x+1) ln(x-2)#

so that, differentiating :

#1/y {dy}/{dx} = ln(x-2) + {x+1}/{x-2} #

so that

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

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Answer 2

Adam Adkins

To find the derivative of y = (x - 2)^(x + 1), you can use the product rule combined with the chain rule. The derivative is given by:

dy/dx = [(x + 1)(x - 2)^(x)] + [(x - 2)^(x + 1) * ln(x - 2)]

So, the derivative of y = (x - 2)^(x + 1) with respect to x is:

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

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.