How do you differentiate #f(x)=(x-x^2sinx)/tanx# using the quotient rule?

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Answer 1 · 2015-12-27T16:58:54

#f'(x)=((1-2xsinx-x^2cosx)tanx-sec^2x(x-x^2sinx))/tan^2x#

If #f(x)=(g(x))/(h(x))#, then

#f'(x)=(g'(x)h(x)-h'(x)g(x))/(h(x))^2#

#g(x)=x-x^2sinx#
#h(x)=tanx#

#g'(x)=1-2xsinx-x^2cosx#
#h'(x)=sec^2x#

Note the use of the product rule to find #g'(x)#.

Thus,

#f'(x)=((1-2xsinx-x^2cosx)tanx-sec^2x(x-x^2sinx))/tan^2x#