Question

Find The derivative of:(cosx/1+sinx)^3​

Answer 1

`-(3 \cdot cos^2x)/((1+sinx)^3)`

Explanation:

`y = ((cosx)/(1+sinx))^3\\\\(dy)/(dx) = 3 \cdot ((cosx)/(1+sinx))^2 \cdot (dy)/(dx)((cosx)/(1+sinx))`
`[ y = x^n\ \ \ => \ \ \ (dy)/(dx) = b \cdot x^(n-1) \ ]`

`= 3 \cdot ((cosx)/(1+sinx))^2 \cdot ((1+sin x(-sinx) - cosx(cosx))/((1+sinx)^2)\\\\`
`[\ (u)/(v) = (v \dcot u'- u \cdot v')/(v^2)\ ]`

`= 3 \cdot ((cosx)/(1+sinx))^2 \cdot (-sin x-sin^2x- cos^2x)/((1+sinx)^2)\\\\= 3 \cdot ((cosx)/(1+sinx))^2 \cdot (-sin x- (sin^2x+ cos^2x))/((1+sinx)^2)\\\\= 3 \cdot ((cosx)/(1+sinx))^2 \cdot (-sin x-1)/((1+sinx)^2)\\\\= 3 \cdot ((cosx)/(1+sinx))^2 \cdot (-1 \cdot(sin x+1))/((1+sinx)^2)\\\\= 3 \cdot ((cosx)/(1+sinx))^2 \cdot (-1)/((1+sinx))\\\\`

`= -3 \cdot (cos^2x)/((1+sinx)^3)`