How do you find the derivative of #y=(1/3) ((x^2)+2)^(3/2)#?

1 Answer
Nov 27, 2016

Answer:

#dy/dx=xsqrt(x^2+2)#

Explanation:

Rewriting slightly:

#y=1/3(x^2+2)^(3/2)#

Notice that #(x^2+2)^(3/2)# is a function in the form #u^(3/2)#. The power rule and the chain tells us that we differentiate this normally, in that the derivative of #x^(3/2)# is #3/2x^(1/2)#, but when there's a function in the middle we will multiply by the derivative of the inner function. Thus, the derivative of #u^(3/2)# is #3/2u^(1/2)*d/dx(u)#.

So, we see that:

#dy/dx=1/3(3/2(x^2+2)^(1/2))*d/dx(x^2+2)#

The power rule tells us that the derivative of #x^2+2# is #2x#:

#dy/dx=1/3(3/2)(2x)(x^2+2)^(1/2)#

Simplifying:

#dy/dx=xsqrt(x^2+2)#