If ` y = cos ^(-1) [sin (4^(x))], " find " (dy)/(dx)`

Answer

To solve the problem of finding \(\frac{dy}{dx}\) for the function \(y = \cos^{-1}[\sin(4^x)]\), we can follow these steps:

### Step 1: Rewrite the Function
We start with the function:
\[ y = \cos^{-1}[\sin(4^x)] \]
We can use the identity that relates sine and cosine:
\[ \sin(4^x) = \cos\left(\frac{\pi}{2} - 4^x\right) \]
Thus, we can rewrite \(y\) as:
\[ y = \cos^{-1}\left[\cos\left(\frac{\pi}{2} - 4^x\right)\right] \]

### Step 2: Simplify the Expression
Since \(\cos^{-1}(\cos(\theta)) = \theta\) when \(\theta\) is in the range of \(\cos^{-1}\), we can simplify:
\[ y = \frac{\pi}{2} - 4^x \]

### Step 3: Differentiate the Function
Now, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2} - 4^x\right) \]
The derivative of a constant (\(\frac{\pi}{2}\)) is 0, so we only need to differentiate \(-4^x\):
\[ \frac{dy}{dx} = -\frac{d}{dx}(4^x) \]
Using the derivative formula for exponential functions, \(\frac{d}{dx}(a^x) = a^x \ln(a)\), we find:
\[ \frac{dy}{dx} = -4^x \ln(4) \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = -4^x \ln(4) \]

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