How do you differentiate #f(x)=sqrtsin(1/lnx^2)# using the chain rule?

1 Answer
Nov 17, 2015

So, let's break this down into a chain of functions:

# f(u) = sqrt(u)#

# u(v) = sin(v)#

# v(w) = 1/ w#

# w(z) = ln z#

# z(x) = x^2#

Now, we need to differentiate every one of those functions:

# f(u) = sqrt(u) = u^(1/2) color(white)(xxx) => color(white)(xx) f'(u) = 1/2 u ^(-1/2) = 1/ (2 sqrt(u)#

# u(v) = sin(v) color(white)(xxxxxii) =>color(white)(xx) u'(v) = cos(v)#

# v(w) = 1/ w color(white)(xxxxxxxii) =>color(white)(xx) v'(w) = -1/w^2#

# w(z) = ln z color(white)(xxxxxxi) =>color(white)(xx) w'(z) = 1/z#

# z(x) = x^2 color(white)(xxxxxxxx) =>color(white)(xx) z'(x) = 2x#

So, the derivative of #f(x)# is the product of all that derivatives:

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

# color(white)(xxxiii) = 1/(2 sqrt( sin( 1 / (ln x^2)))) * cos( 1 / (ln x^2)) * ( - 1 / ln^2 x^2) * 1 / x^2 * 2x#

# color(white)(xxxiii) = - ( cos( 1 / (ln x^2)) )/( sqrt( sin( 1 / (ln x^2))) * ln^2 x^2 * x)#