Question

Differentiate In(cosx+sinx/cosx-sinx)

Answer 1

To differentiate the function \( y = \ln\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right) \), we will apply the chain rule and the quotient rule.

1. **Differentiate using the chain rule**:
The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).

Let \( u = \frac{\cos x + \sin x}{\cos x - \sin x} \).
Then, we find \( \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \).

2. **Find \( \frac{du}{dx} \)** using the quotient rule:
The quotient rule states that if \( u = \frac{f(x)}{g(x)} \), then:
\[
\frac{du}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}
\]
Here, \( f(x) = \cos x + \sin x \) and \( g(x) = \cos x - \sin x \).

- Calculate \( f'(x) \) and \( g'(x) \):
\[
f'(x) = -\sin x + \cos x
\]
\[
g'(x) = -\sin x - \cos x
\]

Substitute into the quotient rule:
\[
\frac{du}{dx} = \frac{(-\sin x + \cos x)(\cos x - \sin x) - (\cos x + \sin x)(-\sin x - \cos x)}{(\cos x - \sin x)^2}
\]

Now, simplify the numerator:
- First term:
\((- \sin x + \cos x)(\cos x - \sin x) = -\sin x \cos x + \sin^2 x + \cos^2 x - \cos x \sin x = \sin^2 x + \cos^2 x - 2sin x cos x = 1 - 2sin x cos x\)

- Second term:
\((\cos x + \sin x)(-\sin x - \cos x) = -\cos x \sin x - \cos^2 x - \sin^2 x - \sin x \cos x\)
= \(-2 \sin x \cos x - 1 \).

Combine both terms:
\[
\frac{du}{dx} = \frac{(1 - 2\sin x \cos x) - (-1 - 2 \sin x \cos x)}{(\cos x - \sin x)^2}
= \frac{2}{(\cos x - \sin x)^2}
\]

3. **Combine results**:
We have
\[
\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{\frac{\cos x + \sin x}{\cos x - \sin x}} \cdot \frac{2}{(\cos x - \sin x)^2} = \frac{2 (\cos x - \sin x)}{(\cos x + \sin x) (\cos x - \sin x)^2}
= \frac{2}{(\cos x + \sin x)(\cos x - \sin x)}
\]

Thus, the derivative of \( \ln\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right) \) is:
\[
\frac{2}{(\cos x + \sin x)(\cos x - \sin x)}
\]