How do you differentiate #y = ln [x^4 sin^2 (x)]#?

1 Answer
Feb 20, 2016

Answer:

#d/dx ln[x^4sin^2(x)]=4/(x)+2 cot(x)#

See explanation for details.

Explanation:

I assume that you know how to use the chain rule and the product rule. If you don't please ask for further explanation in the comments.

Define: #f(x)= ln[x^4sin^2(x)]#

First use the chain rule:
#f'(x)=(1/(x^4sin^2(x))) (d/dx(x^4sin^2(x)))#
Use the product rule to find #d/dx(x^4sin^2(x))#
To do this we'll need to know #d/dx sin^2(x)#:

#d/dx sin^2(x)= d/dx sin(x)sin(x)#
Use the product rule:
#d/dx sin(x)sin(x)=sin(x)cos(x)+cos(x)sin(x)=2sin(x)cos(x)#

Using this result:
#f'(x)=(1/(x^4sin^2(x))) (4x^3sin^2(x)+x^4(2sin(x)cos(x)))#

#=(1/(x^4sin^2(x))) (4x^3)(sin^2(x)+x/2 sin(x)cos(x))#
#=(4/(xsin^2(x))) (sin^2(x)+x/2 sin(x)cos(x))#
#=(4/(x)) (sin^2(x)/sin^2(x)+x/2(sin(x)cos(x))/(sin^2(x)) )#
#=(4/(x)) (sin^2(x)/sin^2(x)+x/2 (sin(x)cos(x))/(sin(x)sin(x)))#
#=(4/(x)) (1+x/2 cos(x)/sin(x)))#
Note that, #cot(x)=1/tan(x)=cos(x)/sin(x)#, so:
#f'(x)=(4/(x)) (1+x/2 cot(x))#

Finally we have:

#d/dx ln[x^4sin^2(x)]=4/(x)+2 cot(x)#