What is the derivative of (x^2)(sinx)(tanx) without using the chain rule?

2 Answers
Apr 13, 2015

You can manipulate your function remembering that:
#tan(x)=sin(x)/cos(x)#
and #sin^2(x)=1-cos^2(x)#
and get:
enter image source here
from this point onwards it is a piece of cake!!!
:-)

Apr 13, 2015

Use the product rule for three factors:

#d/(dx)(fgh) = f'gh+fg'h+fgh'#

For #y = x^2sinxtanx#, we get

#y' = 2xsinxtanx + x^2 cosxtanx+x^2sinxsec^2x#

We ca rewrite the middle term more simply, and we may choose to rewrite the third term:

#y' = 2xsinxtanx + x^2 sinx+x^2tanxsecx#.

#d/(dx)(fgh) =f'[gh]+f d/(dx)[gh]#

#color(white)"sssssssss"# #=f'gh+f[g'h+gh']#

#color(white)"sssssssss"# #=f'gh+fg'h+fgh'#.