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

1 Answer
Nov 26, 2015

Answer:

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

Explanation:

#f(x) = sin ((x/(x^2-1))^(3/2))#

Ok, so let's break your function down into a chain of functions:

#f(u) = sin(u)#

#u(v) = v^(3/2)#

#v(x) = x/(x^2-1)#

As next, you need to differentiate those three functions.

#f(u) = sin(u) color(white)(xxx) => f'(u) = cos(u)#

#u(v) = v^(3/2) color(white)(xxxxxx) => u'(v) = 3/2 v^(1/2)#

To differentiate #v(x)#, you can use the quotient rule:

if #f(x) = g(x) / (h(x))#, the derivative is #f'(x) = (g'(x) * h(x) - h'(x) * g(x)) / (h^2(x))#.

So, in your case, the derivative is:

#v(x) = x/(x^2-1) color(white)(x) => v'(x) = (1 * (x^2-1) - 2x * x)/(x^2-1)^2 = (-x^2 - 1)/(x^2-1)^2#

Now, to compute the derivative of #f(x)#, you just need to multiply your three derivatives! :-)

#f'(x) = f'(u(v(x))) * u'(v(x)) * v'(x) #

# color(white)(xxxx) = cos((x/(x^2-1))^(3/2)) * 3/2 (x/(x^2-1))^(1/2) * (-x^2 - 1)/(x^2-1)^2#

# color(white)(xxxx) = 3/2 cos((x/(x^2-1))^(3/2)) (x/(x^2-1))^(1/2) * (-x^2 - 1)/(x^2-1)^2#