If `y=sin^(-1)(cos x)+cos^(-1)(sin x)`, prove that `(dy)/(dx)=-2`

Options

A. `-2`
B. `2`
C. `1`
D. `0`

Answer

To prove that \(\frac{dy}{dx} = -2\) for the given function \(y = \sin^{-1}(\cos x) + \cos^{-1}(\sin x)\), we can follow these steps:

### Step 1: Rewrite the function
We start with the given function:
\[ y = \sin^{-1}(\cos x) + \cos^{-1}(\sin x) \]

### Step 2: Use the identity for inverse functions
We know that:
\[ \sin^{-1}(a) + \cos^{-1}(a) = \frac{\pi}{2} \]
for any \(a\) in the domain. However, we can also express \(\cos^{-1}(\sin x)\) in terms of \(\sin^{-1}(\cos x)\) as follows:
\[ \cos^{-1}(\sin x) = \frac{\pi}{2} - \sin^{-1}(\sin x) \]
Thus, we can rewrite \(y\) as:
\[ y = \sin^{-1}(\cos x) + \left(\frac{\pi}{2} - \sin^{-1}(\sin x)\right) \]

### Step 3: Simplify the expression
Now, substituting this back into our equation for \(y\):
\[ y = \sin^{-1}(\cos x) + \frac{\pi}{2} - \sin^{-1}(\sin x) \]
Since \(\sin^{-1}(\sin x) = x\) (for \(x\) in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\)), we can simplify further:
\[ y = \sin^{-1}(\cos x) + \frac{\pi}{2} - x \]

### Step 4: Differentiate \(y\) with respect to \(x\)
Now, we differentiate \(y\):
\[ \frac{dy}{dx} = \frac{d}{dx} \left(\sin^{-1}(\cos x) + \frac{\pi}{2} - x\right) \]
The derivative of \(\frac{\pi}{2}\) is \(0\), and the derivative of \(-x\) is \(-1\). For \(\sin^{-1}(\cos x)\), we use the chain rule:
\[ \frac{d}{dx}(\sin^{-1}(\cos x)) = \frac{1}{\sqrt{1 - (\cos x)^2}} \cdot (-\sin x) = -\frac{\sin x}{\sqrt{1 - \cos^2 x}} = -\frac{\sin x}{\sin x} = -1 \]
Thus, we have:
\[ \frac{dy}{dx} = -1 - 1 = -2 \]

### Conclusion
Therefore, we have proved that:
\[ \frac{dy}{dx} = -2 \]