Question

Find the derivative of $\frac{cos\,x}{1+sin\,x}$.

• A. $-(1 + sinx)^2$
• B. $\frac{1}{\left(1+sin\,x\right)}$
• C. $\left(\frac{-1}{1+sin\,x}\right)$
• D. None of these

Answer 1

The Correct Option is C

Let $y=\frac{cos\,x}{1+sin\,x}$ Differentiating both sides with respect to $x$, we get $\frac{dy}{dx}=\frac{d}{dx}\left(\frac{cos\,x}{1+sin\,x}\right)$ $=\frac{\left(1+sin\,x\right) \frac{d}{dx}\left(cos\,x\right)-cos\,x \frac{d}{dx}\left(1+sin\,x\right)}{\left(1+sin\,x\right)^{2}}$ $=\frac{\left(1+sin\,x\right)\left(-sin\,x\right)-cos\,x\left(cos\,x\right)}{\left(1+sin\,x\right)^{2}}$ $=\frac{-sin\,x-sin^{2}\,x-cos^{2}\,x}{\left(1+sin\,x\right)^{2}}$ $=\frac{-\left(1+sin\,x\right)}{\left(1+sin\,x\right)^{2}}=\frac{-1}{1+sin\,x}$