How do you differentiate #f(x)=(x^3-2x+3)^(3/2)# using the chain rule?

1 Answer
Apr 11, 2018

Answer:

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

Explanation:

The chain rule:
#d/dx f(g(x)) = f'(g(x)) * g'(x)#

The power rule:
#d/dx x^n = n*x^(n-1)#

Applying these rules:
1 The inner function, #g(x)# is #x^3-2x+3#, the outer function, #f(x)# is #g(x)^(3/2)#

2 Take the derivative of the outer function using the power rule
#d/dx (g(x))^(3/2) = 3/2 * g(x)^(3/2 - 2/2) = 3/2 * g(x)^(1/2) = 3/2 * sqrt(g(x))#
#f'(g(x)) = 3/2 * sqrt(x^3 - 2x + 3)#

3 Take the derivative of the inner function
#d/dx g(x) = 3x^2 -2 #
#g'(x) = 3x^2 -2 #

4 Multiply #f'(g(x))# with #g'(x)#
# (3/2 * sqrt(x^3 - 2x + 3) ) * (3x^2 - 2)#

solution: #3/2 * (sqrt(x^3 - 2x + 3)) * (3x^2 - 2)#