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

Question

1929 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,
$frac{2x^3sin ^2(x)cot (x)+3x^2sin ^2(x)+x^4}{sin ^2(x)(xcot (x)+1)^2}$

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