Understanding the Derivative Problem

The question asks us to find the derivative of one function with respect to another function. This is a common type of differentiation problem that can be solved using the chain rule. Specifically, if we want to find the derivative of a function \(u\) with respect to a function \(v\), where both \(u\) and \(v\) are functions of \(x\), we can use the formula:

In this problem, the first function is \(u = \ln(x + \sin x)\) and the second function is \(v = x + \cos x\). We need to calculate the derivatives of \(u\) and \(v\) with respect to \(x\) and then divide them.

Step-by-Step Solution to Finding the Derivative

Step 1: Find the derivative of the first function with respect to \(x\).

Let \(u = \ln(x + \sin x)\). To find \(\frac{du}{dx}\), we use the chain rule for the natural logarithm function. The derivative of \(\ln(f(x))\) is \(\frac{f'(x)}{f(x)}\). Here, \(f(x) = x + \sin x\). The derivative of \(f(x)\) with respect to \(x\) is:

So, the derivative of \(u\) with respect to \(x\) is:

Step 2: Find the derivative of the second function with respect to \(x\).

Let \(v = x + \cos x\). To find \(\frac{dv}{dx}\), we differentiate term by term:

So, the derivative of \(v\) with respect to \(x\) is:

Step 3: Divide the derivative of the first function by the derivative of the second function.

Now we apply the formula \(\frac{du}{dv} = \frac{du/dx}{dv/dx}\):

To simplify this complex fraction, we can rewrite it as:

Comparing with the Options

The calculated derivative of \(\ln(x + \sin x)\) with respect to \((x + \cos x)\) is \(\frac{1 + \cos x}{(x + \sin x)(1 - \sin x)}\). Let's look at the given options:

• Option 1: \(\frac{{1 + \cos {\rm{x}}}}{{\left( {{\rm{x}} + \sin {\rm{x}}} \right)\left( {1 - \sin {\rm{x}}} \right)}}\)
• Option 2: \(\frac{{1 - \cos {\rm{x}}}}{{\left( {{\rm{x}} + \sin {\rm{x}}} \right)\left( {1 + \sin {\rm{x}}} \right)}}\)
• Option 3: \(\frac{{1 - \cos {\rm{x}}}}{{\left( {{\rm{x}} - \sin {\rm{x}}} \right)\left( {1 + \cos {\rm{x}}} \right)}}\)
• Option 4: \(\frac{{1 + \cos {\rm{x}}}}{{\left( {{\rm{x}} - \sin {\rm{x}}} \right)\left( {1 - \cos {\rm{x}}} \right)}}\)

Our calculated result matches Option 1 exactly.

Conclusion

The derivative of \(\ln(x + \sin x)\) with respect to \((x + \cos x)\) is found by taking the derivative of \(\ln(x + \sin x)\) with respect to \(x\) and dividing it by the derivative of \((x + \cos x)\) with respect to \(x\). Following the steps using the chain rule and basic differentiation rules leads to the final answer.

Revision Table - Calculus Derivatives

Additional Information - Chain Rule in Differentiation

The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. A composite function is a function that is inside another function, like \(f(g(x))\). The chain rule states that the derivative of \(f(g(x))\) with respect to \(x\) is \(f'(g(x)) \cdot g'(x)\).

In our problem:

• For \(u = \ln(x + \sin x)\), the outer function is \(\ln(y)\) and the inner function is \(y = x + \sin x\). The derivative of the outer function is \(\frac{d}{dy}(\ln y) = \frac{1}{y}\). The derivative of the inner function is \(\frac{dy}{dx} = \frac{d}{dx}(x + \sin x) = 1 + \cos x\). Applying the chain rule, \(\frac{du}{dx} = \frac{1}{y} \cdot (1 + \cos x) = \frac{1}{x + \sin x} \cdot (1 + \cos x) = \frac{1 + \cos x}{x + \sin x}\).
• When finding the derivative of one function with respect to another, like \(\frac{du}{dv}\), we treat both \(u\) and \(v\) as functions of an intermediate variable (in this case, \(x\)) and use the relationship \(\frac{du}{dv} = \frac{du/dx}{dv/dx}\). This is a direct application derived from the chain rule where \(v\) is considered the independent variable for \(u\).