How do you find the derivative of #tanx/(1+sinx)#?

1 Answer
Jim G.
Jul 3, 2017

#((1+sinx)sec^2x-sinx)/(1+sinx)^2#

Explanation:

#"differentiate using the "color(blue)"quotient rule"#

#"given " f(x)=(g(x))/(h(x))" then"#

#f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"#

#g(x)=tanxrArrg'(x)=sec^2x#

#h(x)=1+sinxrarrh'(x)=cosx#

#rArrf'(x)=((1+sinx).sec^2x-tanxcosx)/(1+sinx)^2#

#color(white)(rArrf'(x))=((1+sinx)sec^2x-sinx)/(1+sinx)^2#

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