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

Hint:
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 \[\dfrac{{d\left[ {\ln \left( {f\left( x \right)} \right)} \right]}}{{dx}} \times \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} \times \dfrac{{d\left[ x \right]}}{{dx}}\].