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

1 Answer
Noah G
Feb 15, 2017

#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!

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