Differentiate the following w.r.t. x :
`sqrt(tanx) a^(x)`.

Answer

To differentiate the function \( y = \sqrt{\tan x} \cdot a^x \) with respect to \( x \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

In our case, we can identify:
- \( u = \sqrt{\tan x} \)
- \( v = a^x \)

### Step 1: Differentiate \( u \) and \( v \)

1. **Differentiate \( u = \sqrt{\tan x} \)**:
- We can use the chain rule. The derivative of \( \sqrt{f(x)} \) is \( \frac{1}{2\sqrt{f(x)}} \cdot f'(x) \).
- Here, \( f(x) = \tan x \), so \( f'(x) = \sec^2 x \).
- Therefore,
\[ \frac{du}{dx} = \frac{1}{2\sqrt{\tan x}} \cdot \sec^2 x \]

2. **Differentiate \( v = a^x \)**:
- The derivative of \( a^x \) is given by:
\[ \frac{dv}{dx} = a^x \ln a \]

### Step 2: Apply the Product Rule

Now, applying the product rule:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Substituting the values we found:

\[ \frac{dy}{dx} = \sqrt{\tan x} \cdot (a^x \ln a) + a^x \cdot \left(\frac{1}{2\sqrt{\tan x}} \cdot \sec^2 x\right) \]

### Step 3: Simplify the Expression

Combining the terms, we have:

\[ \frac{dy}{dx} = a^x \ln a \sqrt{\tan x} + \frac{a^x \sec^2 x}{2\sqrt{\tan x}} \]

### Final Answer

Thus, the derivative of \( y = \sqrt{\tan x} \cdot a^x \) with respect to \( x \) is:

\[ \frac{dy}{dx} = a^x \ln a \sqrt{\tan x} + \frac{a^x \sec^2 x}{2\sqrt{\tan x}} \]

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