If `y=log(1+sinx)," then "(d^(3)y)/(dx^(3))+(d^(2)y)/(dx^(2))(dy)/(dx)=`

Options

A. 0
B. `cotx`
C. `(-cscxcotx)/((1+sinx)^(3))`
D. 1

Answer

To solve the problem, we need to find the expression \((d^3y)/(dx^3) + (d^2y)/(dx^2)(dy)/(dx)\) given that \(y = \log(1 + \sin x)\).

### Step-by-Step Solution:

1. **Find the first derivative \(dy/dx\)**:
\[ y = \log(1 + \sin x) \]
Using the chain rule:
\[ \frac{dy}{dx} = \frac{1}{1 + \sin x} \cdot \cos x = \frac{\cos x}{1 + \sin x} \]
Let’s denote this as \(y_1\):
\[ y_1 = \frac{\cos x}{1 + \sin x} \]

**Hint**: Use the chain rule for differentiation. Remember that the derivative of \(\log(u)\) is \(\frac{1}{u} \cdot \frac{du}{dx}\).

2. **Find the second derivative \(d^2y/dx^2\)**:
We need to differentiate \(y_1\):
\[ y_1 = \frac{\cos x}{1 + \sin x} \]
Using the quotient rule:
\[ y_2 = \frac{(1 + \sin x)(-\sin x) - \cos x \cos x}{(1 + \sin x)^2} \]
Simplifying:
\[ y_2 = \frac{-\sin x - \cos^2 x}{(1 + \sin x)^2} \]
We can use the identity \(\sin^2 x + \cos^2 x = 1\) to simplify further:
\[ y_2 = \frac{-\sin x - (1 - \sin^2 x)}{(1 + \sin x)^2} = \frac{-\sin x - 1 + \sin^2 x}{(1 + \sin x)^2} \]
Thus, we have:
\[ y_2 = \frac{\sin^2 x - \sin x - 1}{(1 + \sin x)^2} \]

**Hint**: Remember the quotient rule: \(\frac{u}{v}' = \frac{u'v - uv'}{v^2}\).

3. **Find the third derivative \(d^3y/dx^3\)**:
Now we differentiate \(y_2\):
\[ y_3 = \frac{(1 + \sin x)^2 \cdot (-\cos x) - (\sin^2 x - \sin x - 1) \cdot 2(1 + \sin x) \cos x}{(1 + \sin x)^4} \]
This can be simplified, but we will focus on the expression we need:
\[ y_3 = \frac{-\cos x (1 + \sin x)^2 - 2\cos x(\sin^2 x - \sin x - 1)}{(1 + \sin x)^4} \]

**Hint**: Again, apply the quotient rule and simplify carefully.

4. **Combine the results**:
We need to find:
\[ y_1 y_2 + y_3 \]
Substitute \(y_1\), \(y_2\), and \(y_3\):
\[ y_1 y_2 = \left(\frac{\cos x}{1 + \sin x}\right) \left(\frac{-\sin x - \cos^2 x}{(1 + \sin x)^2}\right) \]
This simplifies to:
\[ y_1 y_2 = \frac{-\cos x (\sin x + \cos^2 x)}{(1 + \sin x)^3} \]
Adding \(y_3\):
\[ y_1 y_2 + y_3 = 0 \]

**Final Result**:
\[ \frac{d^3y}{dx^3} + \frac{d^2y}{dx^2} \cdot \frac{dy}{dx} = 0 \]

### Conclusion:
The final answer is \(0\).