Question

Diffrentiate 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 can use the chain rule and the quotient rule.

First, let's differentiate it step-by-step:

1. **Differentiate the outer function**:
\[
\frac{dy}{dx} = \frac{1}{\frac{\cos x + \sin x}{\cos x - \sin x}} \cdot \frac{d}{dx} \left(\frac{\cos x + \sin x}{\cos x - \sin x}\right)
\]

2. **Differentiate the inner function** using the quotient rule:
If we let \( u = \cos x + \sin x \) and \( v = \cos x - \sin x \), then by the quotient rule,
\[
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]

- Compute \( \frac{du}{dx} \):
\[
\frac{du}{dx} = -\sin x + \cos x = \cos x - \sin x
\]

- Compute \( \frac{dv}{dx} \):
\[
\frac{dv}{dx} = -\sin x - \cos x = -(\sin x + \cos x)
\]

Now, substitute these into the quotient rule:
\[
\frac{d}{dx}\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right) = \frac{(\cos x - \sin x)(\cos x - \sin x) - (\cos x + \sin x)(-\sin x - \cos x)}{(\cos x - \sin x)^2}
\]

Simplifying the numerator:
- First term: \( (\cos x - \sin x)^2 = \cos^2 x - 2\cos x \sin x + \sin^2 x = 1 - 2\sin x \cos x = 1 - \sin(2x) \)
- Second term: \((\cos x + \sin x)(-\sin x - \cos x) = -(\cos^2 x +\sin^2 x + 2\sin x \cos x) = -1 - \sin(2x)\)

Thus, we have:
\[
\frac{d}{dx}\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right) = \frac{(1 - \sin(2x)) - (-1 - \sin(2x))}{(\cos x - \sin x)^2} = \frac{2}{(\cos x - \sin x)^2}
\]

3. **Combine the results**:
Putting it all together, we have:
\[
\frac{dy}{dx} = \frac{1}{\frac{\cos x + \sin x}{\cos x - \sin x}} \cdot \frac{2}{(\cos x - \sin x)^2}
\]
This can be rewritten as:
\[
\frac{dy}{dx} = \frac{2(\cos x - \sin x)}{(\cos x + \sin x)(\cos x - \sin x)^2} = \frac{2}{(\cos x + \sin x)(\cos x - \sin x)}
\]

Finally, the derivative is:
\[
\frac{dy}{dx} = \frac{2}{(\cos x + \sin x)(\cos x - \sin x)}.
\]