How do you find the derivative of #y=x^5/(4sinx)#?

1 Answer
Jim G.
Apr 8, 2018

#dy/dx=(x^4(5sinx-xcosx))/(4sin^2x)#

Explanation:

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

#"Given "y=(g(x))/(h(x))" then"#

#dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#

#g(x)=x^5rArrg'(x)=5x^4#

#h(x)=4sinxrArrh'(x)=4cosx#

#rArrdy/dx=(20x^4sinx-4x^5cosx)/(4sinx)^2#

#color(white)(rArrdy/dx)=(4x^4(5sinx-xcosx))/(16sin^2x)#

#color(white)(rArrdy/dx)=(x^4(5sinx-xcosx))/(4sin^2x)#

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