Question

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

Answer 1

Answer:

$-\frac{3 \cdot cos^2x}{(1+sinx)^3}$

Step-by-step explanation:

$y = (\frac{cosx}{1+sinx})^3\\\\\frac{dy}{dx} = 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{dy}{dx}(\frac{cosx}{1+sinx})$                        $[ y = x^n\ \ \ => \ \ \ \frac{dy}{dx} = b \cdot x^{n-1} \ ]$

    $= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{(1+sin x(-sinx) - cosx(cosx)}{(1+sinx)^2}$     $[\ \frac{u}{v} = \frac{v \dcot u'- u \cdot v'}{v^2}\ ]$

    $= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-sin x-sin^2x- cos^2x}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-sin x- (sin^2x+ cos^2x)}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-sin x-1}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-1 \cdot(sin x+1)}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-1}{(1+sinx)}$

   $= -3 \cdot \frac{cos^2x}{(1+sinx)^3}$