Question

differentiation by quotient rule :
y = cot / 1+ cot

Answer 1

$$\begin{gathered} \mathrm{Let}y=\frac{cotx}{1+cotx} \\ \Rightarrow \frac{dy}{dx}=\frac{\left(1+cotx\right)\times \left(-\cos e{c}^{2}x\right)-cotx\left(-\cos ec{}^{2}x\right)}{{\left(1+cotx\right)}^2} \\ \Rightarrow \frac{dy}{dx}=\frac{-\cos e{c}^{2}x-cotx.\cos e{c}^{2}x+cotx.\cos e{c}^{2}x}{{\left(1+cotx\right)}^2} \\ \Rightarrow \frac{dy}{dx}=\frac{-\cos e{c}^{2}x}{{\left(1+cotx\right)}^2} \end{gathered}$$