The differential cofficient of `a^(sin^(-1)x)w.r.t sin^(-1)x` is -

Options

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

Answer

To find the differential coefficient of \( a^{\sin^{-1} x} \) with respect to \( \sin^{-1} x \), we can follow these steps:

### Step 1: Define the Functions
Let:
- \( f(x) = a^{\sin^{-1} x} \)
- \( g(x) = \sin^{-1} x \)

We want to find \( \frac{f'(x)}{g'(x)} \).

### Step 2: Differentiate \( f(x) \)
To differentiate \( f(x) = a^{\sin^{-1} x} \), we can use logarithmic differentiation. Taking the natural logarithm of both sides gives:

\[ \ln f(x) = \sin^{-1} x \cdot \ln a \]

Now, differentiate both sides with respect to \( x \):

\[ \frac{1}{f(x)} f'(x) = \frac{d}{dx}(\sin^{-1} x) \cdot \ln a \]

Using the derivative of \( \sin^{-1} x \):

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

So we have:

\[ \frac{1}{f(x)} f'(x) = \frac{\ln a}{\sqrt{1 - x^2}} \]

Multiplying both sides by \( f(x) \):

\[ f'(x) = a^{\sin^{-1} x} \cdot \frac{\ln a}{\sqrt{1 - x^2}} \]

### Step 3: Differentiate \( g(x) \)
Now, differentiate \( g(x) = \sin^{-1} x \):

\[ g'(x) = \frac{1}{\sqrt{1 - x^2}} \]

### Step 4: Find \( \frac{f'(x)}{g'(x)} \)
Now we can find the ratio of the derivatives:

\[ \frac{f'(x)}{g'(x)} = \frac{a^{\sin^{-1} x} \cdot \frac{\ln a}{\sqrt{1 - x^2}}}{\frac{1}{\sqrt{1 - x^2}}} \]

The \( \sqrt{1 - x^2} \) terms cancel out:

\[ \frac{f'(x)}{g'(x)} = a^{\sin^{-1} x} \cdot \ln a \]

### Final Result
Thus, the differential coefficient of \( a^{\sin^{-1} x} \) with respect to \( \sin^{-1} x \) is:

\[ \boxed{a^{\sin^{-1} x} \cdot \ln a} \]