How do you differentiate #f(x) = x^3/(xcotx+1)# using the quotient rule?

Question

1634 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-21T04:56:52

#frac{d}{dx}(frac{x^3}{xcot (x)+1})=frac{2x^3sin ^2(x)cot (x)+3x^2sin ^2(x)+x^4}{sin ^2(x)(xcot (x)+1)^2}#

#frac{d}{dx}(frac{x^3}{xcot (x)+1})#

#(frac{f}{g})^'=frac{f^'cdot g-g^'cdot f}{g^2}#

#=frac{frac{d}{dx}(x^3)(xcot (x)+1)-frac{d}{dx}(xcot (x)+1)x^3}{(xcot (x)+1)^2}#

We know,
#frac{d}{dx}(x^3)=3x^2#
#frac{d}{dx}cot (x)+1=-frac{x}{sin ^2(x)}+cot(x)#

so,#=frac{3x^2(xcot (x)+1)-(-frac{x}{sin ^2(x)}+cot (x))x^3}{(xcot (x)+1)^2}#

Simplifying it,
#frac{2x^3sin ^2(x)cot (x)+3x^2sin ^2(x)+x^4}{sin ^2(x)(xcot (x)+1)^2}#