If #f(x) =csc^3(x/2) # and #g(x) = sqrt(2x+3 #, what is #f'(g(x)) #?

1 Answer
Apr 28, 2017

Answer:

#color(red){f'(g(x)) = -3/[2sqrt(2x+3)]*csc^3(sqrt(2x+3)/2)*cot(sqrt(2x+3)/2)}#

Explanation:

#f(x) = csc^3(x/2)#

#g(x) = sqrt(2x+3)#

#therefore f(g(x)) = csc^3(sqrt(2x+3)/2)#

#=> f'(x) = [d[csc^3(sqrt(2x+3)/2)]]/dx#

Using chain rule to evaluate #[d[csc^3(sqrt(2x+3)/2)]]/dx#

#=> f'(x) = [d[csc^3(sqrt(2x+3)/2)]]/(d[csc(sqrt(2x+3)/2)])xx (d[csc(sqrt(2x+3)/2)])/(d(sqrt(2x+3)/2)) xx {d(sqrt(2x+3)/2)}/(d(2x+3)) xx [d(2x+3)]/dx#

Now,

1.

#[d[csc^3(sqrt(2x+3)/2)]]/(d[csc(sqrt(2x+3)/2)]) = 3*csc^(3-1)(sqrt(2x+3)/2) =color(green){3csc^2(sqrt(2x+3)/2)#

2.

# (d[csc(sqrt(2x+3)/2)])/(d(sqrt(2x+3)/2)) = color(green){-csc(sqrt(2x+3)/2)*cot(sqrt(2x+3)/2)}#

3.

#{d(sqrt(2x+3)/2)}/(d(2x+3)) = 1/2[d(sqrt(2x+3))]/[d(2x+3)] = 1/2*(d(2x+3)^(1/2))/(d(2x+3)) #

#=1/2*1/2*(2x+3)^(1/2-1) = 1/4*(2x+3)^(-1/2) = color(green)[1/(4sqrt(2x+3))]#

4.

#[d(2x+3)]/dx = (d(2x))/dx+(d(3))/dx# . [Using sum rule]
# = 2dx/dx+0 = color(green)2#

#therefore f'(g(x))=color(green){3csc^2(sqrt(2x+3)/2) xx color(green){-csc(sqrt(2x+3)/2)*cot(sqrt(2x+3)/2)} xx color(green)[1/(4sqrt(2x+3))] xx color(green)2#

#= color(red){-3/[2sqrt(2x+3)]*csc^3(sqrt(2x+3)/2)*cot(sqrt(2x+3)/2)}#

Hence,
#color(red){f'(g(x)) = -3/[2sqrt(2x+3)]*csc^3(sqrt(2x+3)/2)*cot(sqrt(2x+3)/2)}#