How do you differentiate #g(y) =sqrtx (x^2 - 2)^(3/2 # using the product rule?

1 Answer
Mar 4, 2017

Answer:

#g'(x)=(x^2-2)^(3/2)/(2sqrtx)# #+3x^(3/2)sqrt(x^2-2)#

Explanation:

I'm assuming that "#g(y)#" in the question is supposed to be "#g(x)#"
This exercise will require the product rule as well as the chain rule.

Recall that the product rule differentiates two functions that are products of one another.

For the function #h(x)=f(x)*g(x)#,

(1) #h'(x)=f'(x)*g(x)+f(x)*g'(x)#

Recall that the chain rule differentiates a function that is a composition of two functions.

For the function #h(x)=f(g(x))#

(2) #h'(x)=f'(g(x))*g'(x)#

To begin, write all powers as exponents:

#g(x)=x^(1/2)(x^2-2)^(3/2)#

Differentiate

#g'(x)=(1/2)x^(-1/2)*(x^2-2)^(3/2)#

#+x^(1/2)*(3/2)(x^2-2)^(1/2)*2x#

From here, we clean it up.

#g'(x)=(x^2-2)^(3/2)/(2x^(1/2)# #+3x^(3/2)(x^2-2)^(1/2)#

#g'(x)=(x^2-2)^(3/2)/(2sqrtx)# #+3x^(3/2)sqrt(x^2-2)#