The derivative of In(x + sin x) with respect to (x + cos x) is
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:
\[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} \]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:
\[ f'(x) = \frac{d}{dx}(x + \sin x) = \frac{d}{dx}(x) + \frac{d}{dx}(\sin x) = 1 + \cos x \]So, the derivative of \(u\) with respect to \(x\) is:
\[ \frac{du}{dx} = \frac{1 + \cos x}{x + \sin x} \]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:
\[ \frac{dv}{dx} = \frac{d}{dx}(x + \cos x) = \frac{d}{dx}(x) + \frac{d}{dx}(\cos x) = 1 - \sin x \]So, the derivative of \(v\) with respect to \(x\) is:
\[ \frac{dv}{dx} = 1 - \sin x \]Step 3: Divide the derivative of the first function by the derivative of the second function.
Now we apply the formula \(\frac{du}{dv} = \frac{du/dx}{dv/dx}\):
\[ \frac{du}{dv} = \frac{\frac{1 + \cos x}{x + \sin x}}{1 - \sin x} \]To simplify this complex fraction, we can rewrite it as:
\[ \frac{du}{dv} = \frac{1 + \cos x}{x + \sin x} \cdot \frac{1}{1 - \sin x} \] \[ \frac{du}{dv} = \frac{1 + \cos x}{(x + \sin x)(1 - \sin x)} \]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
| Function | Derivative w.r.t. \(x\) |
|---|---|
| \(x^n\) | \(nx^{n-1}\) |
| \(\sin x\) | \(\cos x\) |
| \(\cos x\) | \(-\sin x\) |
| \(\ln x\) | \(\frac{1}{x}\) |
| \(\ln(f(x))\) | \(\frac{f'(x)}{f(x)}\) (Chain Rule) |
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 \(u\).
Similar Questions
What is the derivative of sin(ln x) + cos(ln x) with respect to x at x = e?
Which of the following equations is/are correct?
1. f(-2) = f(5)
2. f”(-2) + f”(0.5) + f”(3) = 4
Select the correct answer using the code given below:
Let f(x) be a polynomial function such that f ∘ f(x) = x 4. What is f'(1) equal to ?
What is the derivative of \(\rm e^{e^x}\) with respect to e x?
What f’(x) equal to when 0 < x < 1?
What is f’(x) equal to when x > 1?
If \(\rm x^m y^n =a^{m+n}\) , then what is \(\rm \dfrac{dy}{dx}\) equal to?
A function is defined in (0, ∞) by \( f(x) = \begin{cases} 1-x^2 & for& , 0 < x \leq 1 \quad \\ In \ x & for &, 1 < x \leq 2 \\ In \ 2 - 1 + 0.5x & for &, 2 < x < \infty \end{cases} \)
Which one of the following is correct in respect of the derivative of the function, i.e. f’(x)?Let f(x + y) = f(x) f(y) for all x and y. Then what is f’(5) equal to [where f’(x) is the derivative of f(x)]?
Which of the following statements are not correct?
1. y as a function of x is not defined for all real x.
2. y as a function of x is not continuous at x = 0
3. y as a function of x is differentiable for all x.
Select the correct answer using the code given below
Important Questions from Evaluation of derivatives
What is the derivative of sin(ln x) + cos(ln x) with respect to x at x = e?
Which of the following equations is/are correct?
1. f(-2) = f(5)
2. f”(-2) + f”(0.5) + f”(3) = 4
Select the correct answer using the code given below:
Let f(x) be a polynomial function such that f ∘ f(x) = x 4. What is f'(1) equal to ?
What is the derivative of \(\rm e^{e^x}\) with respect to e x?
What f’(x) equal to when 0 < x < 1?
