If `y=e^(sin^(-1)x)" and "u=logx," then"(dy)/(du),` is

Options

A. `(e^(sin^(-1)x))/(sqrt(1-x^(2)))`
B. `xe^(sin^(-1)x)`
C. `(xe^(sin^(-1)x))/(sqrt(1-x^(2)))`
D. `(e^(sin^(-1)x))/(x)`

Answer

To find \(\frac{dy}{du}\) when \(y = e^{\sin^{-1} x}\) and \(u = \log x\), we will use the chain rule of differentiation. Here are the steps:

### Step 1: Differentiate \(y\) with respect to \(x\)
We start with the function \(y = e^{\sin^{-1} x}\). To differentiate this, we apply the chain rule.

\[ \frac{dy}{dx} = \frac{d}{dx}(e^{\sin^{-1} x}) = e^{\sin^{-1} x} \cdot \frac{d}{dx}(\sin^{-1} x) \]

### Step 2: Differentiate \(\sin^{-1} x\) with respect to \(x\)
The derivative of \(\sin^{-1} x\) is given by the formula:

\[ \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \]

Substituting this back into our equation for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1 - x^2}} \]

### Step 3: Differentiate \(u\) with respect to \(x\)
Next, we differentiate \(u = \log x\):

\[ \frac{du}{dx} = \frac{d}{dx}(\log x) = \frac{1}{x} \]

### Step 4: Use the chain rule to find \(\frac{dy}{du}\)
Now, we can find \(\frac{dy}{du}\) using the relationship:

\[ \frac{dy}{du} = \frac{dy}{dx} \cdot \frac{dx}{du} \]

Since \(\frac{dx}{du} = \frac{1}{\frac{du}{dx}} = x\), we have:

\[ \frac{dy}{du} = \frac{dy}{dx} \cdot x \]

Substituting \(\frac{dy}{dx}\) from Step 2:

\[ \frac{dy}{du} = \left(e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1 - x^2}}\right) \cdot x \]

### Final Result
Thus, we can express \(\frac{dy}{du}\) as:

\[ \frac{dy}{du} = \frac{x \cdot e^{\sin^{-1} x}}{\sqrt{1 - x^2}} \]