Question

The derivative of $\ln \left( {x + \sin x} \right)$ with respect to $\left( {x + \cos x} \right)$ is
(a) $\dfrac{{1 + \cos x}}{{\left( {x + \sin x} \right)\left( {1 - \sin x} \right)}}$
(b) $\dfrac{{1 - \cos x}}{{\left( {x + \sin x} \right)\left( {1 + \sin x} \right)}}$
(c) $\dfrac{{1 - \cos x}}{{\left( {x - \sin x} \right)\left( {1 + \cos x} \right)}}$
(d) $\dfrac{{1 + \cos x}}{{\left( {x - \sin x} \right)\left( {1 - \cos x} \right)}}$

Answer 1

Here, we need to find the derivative of $\ln \left( {x + \sin x} \right)$ with respect to $\left( {x + \cos x} \right)$. Let $u = \ln \left( {x + \sin x} \right)$ and $v = \left( {x + \cos x} \right)$. We will differentiate these two equations with respect to $x$ to find the value of $\dfrac{{du}}{{dx}}$ and $\dfrac{{dv}}{{dx}}$. Using these values, we can find the value of $\dfrac{{du}}{{dv}}$, and hence, the derivative of $\ln \left( {x + \sin x} \right)$ with respect to $\left( {x + \cos x} \right)$.
Formula Used: The derivative of the sum of two functions is the sum of the derivatives of the two functions, that is $\dfrac{{d\left[ {f\left( x \right) + g\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} + \dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}$.

Complete step by step solution:
Let $u = \ln \left( {x + \sin x} \right)$ and $v = \left( {x + \cos x} \right)$.
We need to find $\dfrac{{d\left[ {\ln \left( {x + \sin x} \right)} \right]}}{{d\left( {x + \cos x} \right)}}$, that is $\dfrac{{du}}{{dv}}$.
We will differentiate the equations $u = \ln \left( {x + \sin x} \right)$ and $v = \left( {x + \cos x} \right)$.
First, we will differentiate both sides of $u = \ln \left( {x + \sin x} \right)$ with respect to $x$.
Differentiating both sides, we get
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{d\left[ {\ln \left( {x + \sin x} \right)} \right]}}{{dx}}$
The derivative of $\ln \left[ {f\left( x \right)} \right]$ is $\dfrac{1}{{f\left( x \right)}} \times \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}$.
Therefore, we get
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{x + \sin x}} \times \dfrac{{d\left( {x + \sin x} \right)}}{{dx}}$
The derivative of the sum of two functions is the sum of the derivatives of the two functions, that is $\dfrac{{d\left[ {f\left( x \right) + g\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} + \dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}$.
Therefore, the equation becomes
$\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{x + \sin x}} \times \left[ {\dfrac{{d\left( x \right)}}{{dx}} + \dfrac{{d\left( {\sin x} \right)}}{{dx}}} \right]$
The derivative of $x$ with respect to $x$ is 1.
The derivative of $\sin x$ with respect to $x$ is $\cos x$.
Therefore, the equation becomes
\Rightarrow \dfrac{{du}}{{dx}} = \dfrac{1}{{x + \sin x}} \times \left[ {1 + \cos x} \right] \\\ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{{1 + \cos x}}{{x + \sin x}} \\\
Now, we will differentiate both sides of $v = \left( {x + \cos x} \right)$ with respect to $x$.
Differentiating both sides, we get
$\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{d\left( {x + \cos x} \right)}}{{dx}}$
Therefore, the equation becomes
$\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{d\left( x \right)}}{{dx}} + \dfrac{{d\left( {\cos x} \right)}}{{dx}}$
The derivate of $x$ with respect to $x$ is 1.
The derivative of $\cos x$ with respect to $x$ is $- \sin x$.
Therefore, the equation becomes
\Rightarrow \dfrac{{dv}}{{dx}} = 1 + \left( { - \sin x} \right) \\\ \Rightarrow \dfrac{{dv}}{{dx}} = 1 - \sin x \\\
Now, we will find the value of $\dfrac{{du}}{{dv}}$.
Multiplying and dividing the expression by $dx$, we get
$\Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{du}}{{dv}} \times \dfrac{{dx}}{{dx}}$
Rewriting the expression, we get
\Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{du}}{{dx}} \times \dfrac{{dx}}{{dv}} \\\ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{du}}{{dx}} \times \dfrac{1}{{\dfrac{{dv}}{{dx}}}} \\\
Substituting $\dfrac{{du}}{{dx}} = \dfrac{{1 + \cos x}}{{x + \sin x}}$ and $\dfrac{{dv}}{{dx}} = 1 - \sin x$ in the equation, we get
\Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{du}}{{dx}} \times \dfrac{{dx}}{{dv}} \\\ \Rightarrow \dfrac{{du}}{{dv}} = \dfrac{{1 + \cos x}}{{x + \sin x}} \times \dfrac{1}{{1 - \sin x}} \\\
Therefore, we get
$\therefore \dfrac{{du}}{{dv}} = \dfrac{{1 + \cos x}}{{\left( {x + \sin x} \right)\left( {1 - \sin x} \right)}}$
Therefore, we get the derivative of $\ln \left( {x + \sin x} \right)$ with respect to $\left( {x + \cos x} \right)$ as $\dfrac{{1 + \cos x}}{{\left( {x + \sin x} \right)\left( {1 - \sin x} \right)}}$.

Thus, the correct option is option (a).

Note:
You should remember to use the chain rule of differentiation. A common mistake is to write the derivative of $\ln \left( {x + \sin x} \right)$ as $\dfrac{1}{{x + \sin x}}$. This is incorrect. According to the chain rule, the derivative of a function $\ln f\left( x \right)$ is given as $\dfrac{{d\left[ {\ln \left( {f\left( x \right)} \right)} \right]}}{{dx}} \times \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} \times \dfrac{{d\left[ x \right]}}{{dx}}$.