Question

The derivative of $ln(x+\sin x)$ with respect to $(x+\cos x)$ is

• A. $\frac{1+\cos x}{(x+\sin x)(1-\sin x)}$
• B. $\frac{1-\cos x}{(x+\sin x)(1+\sin x)}$
• C. $\frac{1-\cos x}{(x-\sin x)(1+\cos x)}$
• D. $\frac{1+\cos x}{(x-\sin x)(1-\cos x)}$

Answer 1

The correct option is A $\frac{1+\cos x}{(x+\sin x)(1-\sin x)}$

Let $u=ln(x+\sin x)$ and $v=x+\cos x$

Now,

$\frac{du}{dx}=\frac{1}{x+\sin x}(1+\cos x)$ .... $\left(i\right)$

and $\frac{dv}{dx}=1-\sin x$ ..... $\left(ii\right)$

Now, we can find derivative of $u$ with respect to $v$ by dividing $\left(i\right)$ by $\left(ii\right)$, we get

$\frac{du/dx}{dv/dx}=\frac{(1+\cos x)}{(x+\sin x)(1-\sin x)}$

or, $\frac{du}{dv}=\frac{1+\cos x}{(x+\sin x)(1-\sin x)}$

Hence, A is the correct option.