How do you differentiate #f(x)=csc(ln(1/x)) # using the chain rule?

1 Answer
Nov 22, 2016

Answer:

#d/dx(f(x))=1/xcotln(1/x)cscln(1/x)#

Explanation:

#f(x)# composed of two functions #cscx " "and" "Ln x#.
#" "#
So, Differentiating #f(x)# is determined by applying chain rule.
#" "#
Let #u(x)=ln(1/x)" " then " "f(x)= csc(u(x))#
#" "#
#d/dx(f(x))=d/dx(cscx(u(x)))#
#" "#
#color(blue)(d/dx(f(x))=u'(x) xx csc'(u(x)))#
#" "#
Let us compute #color(blue)(u'(x))#
#" "#
#u'(x)=(color(red)((1/x)'))/(1/x)#
#" "#
What is the derivative of #1/x=?#
#" "#
The differiation of #1/x# is determined by applying the quotient rule .
#" "#
#color(red)((1/x)')=((1)'xxx-(x)'xx1)/x^2= (0-1)/x^2=color(red)(-1/x^2)#
#" "#
#color(blue)(u'(x))=(-1/x^2)/(1/x)=color(blue)(-1/x)#
#" "#
#" "#
Knowing that the derivative of #cscx" "#is :
#" "#
#color(blue)((cscx)'=-cotxcscx#
#" "#
Let us compute #color(blue)(csc'(u(x))#
#" "#
#color(blue)(csc'(u(x)))=-cot(u(x))xxcsc(u(x))=color(blue)(-cotln(1/x)xxcscln(1/x)#
#" "#
#color(blue)(d/dx(f(x))=u'(x) xx csc'(u(x)))#
#" "#
#d/dx(f(x))=(-1/x)xx(-cotln(1/x)xxcscln(1/x))#
#" "#
Therefore,
#" "#
#d/dx(f(x))=1/xcotln(1/x)cscln(1/x)#