If `f(x)=cosx.g(x)=logx." and "y="gof"(x)` then what is the value of `(dy)/(dx)` at `x=0` ?

Options

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

Answer

To find the value of \(\frac{dy}{dx}\) at \(x=0\) for the function \(y = g(f(x))\) where \(f(x) = \cos x\) and \(g(x) = \ln x\), we can follow these steps:

### Step 1: Define the functions
We have:
- \(f(x) = \cos x\)
- \(g(x) = \ln x\)

### Step 2: Substitute \(f(x)\) into \(g(x)\)
We need to find \(y\):
\[ y = g(f(x)) = g(\cos x) = \ln(\cos x) \]

### Step 3: Differentiate \(y\) with respect to \(x\)
To find \(\frac{dy}{dx}\), we will use the chain rule:
\[ \frac{dy}{dx} = \frac{dg}{df} \cdot \frac{df}{dx} \]
First, we find \(\frac{dg}{df}\) where \(g(f) = \ln(f)\):
\[ \frac{dg}{df} = \frac{1}{f} = \frac{1}{\cos x} \]

Next, we find \(\frac{df}{dx}\):
\[ \frac{df}{dx} = \frac{d}{dx}(\cos x) = -\sin x \]

Now, we can combine these results:
\[ \frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \]

### Step 4: Evaluate \(\frac{dy}{dx}\) at \(x=0\)
Now we need to find \(\frac{dy}{dx}\) at \(x=0\):
\[ \frac{dy}{dx} \bigg|_{x=0} = -\tan(0) = -0 = 0 \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) at \(x=0\) is:
\[ \boxed{0} \]