How do you differentiate #f(x)=sin(1/sqrt(3x^2-4) ) # using the chain rule?

1 Answer
Jan 21, 2016

Answer:

#f'(x) = - cos( 1 / (3x^2 - 4)) * (6x) / (2 sqrt((3x^2-4)^3))#

Explanation:

Let's break down your function using the chain rule:

#f(x) = sin(color(blue)(1/sqrt(3x^2-4)) ) = sin color(blue)(u)#

where #" "u = 1 / sqrt(color(orange)(3x^2-4)) = 1 / sqrt(color(orange)(v)) #

where #" " v = 3x^2 - 4#

According to the chain rule, the derivative is:

#f'(x) = [sin u]' * u' = [sin u]' * [1 / sqrt(v)]' * v' #

Let's compute the derivatives of those terms!

# [sin u]' = cos u = cos( 1 / (3x^2 - 4))#

#u' = [1 / sqrt(v)]' = [v^(-1/2)]' = -1/2 v^(-3/2) = - 1 / (2 sqrt(v^3)) = - 1 / (2 sqrt((3x^2-4)^3))#

#v' = [3x^2 - 4]' = 6x#

Thus, your derivative is:

#f'(x) = [sin u]' * [1 / sqrt(v)]' * v' #

# = cos( 1 / (3x^2 - 4)) * (- 1 / (2 sqrt((3x^2-4)^3))) * 6x#

# = - cos( 1 / (3x^2 - 4)) * (6x) / (2 sqrt((3x^2-4)^3))#