The derivative of `f(x)=cos^(-1)((1)/(sqrt3)(2cosx-3sinx))+{sin^(-1)((1)/(sqrt3)(2cosx+3sinx))}` w.r.t. `sqrt(1+x^(2))` at `x=(1)/(sqrt3)`is.........

Answer

To find the derivative of the function

\[ f(x) = \cos^{-1}\left(\frac{1}{\sqrt{3}}(2\cos x - 3\sin x)\right) + \sin^{-1}\left(\frac{1}{\sqrt{3}}(2\cos x + 3\sin x)\right) \]

with respect to \( \sqrt{1 + x^2} \) at \( x = \frac{1}{\sqrt{3}} \), we will follow these steps:

### Step 1: Simplifying the Function

First, we simplify the arguments of the inverse trigonometric functions.

Let:
\[ A = \frac{1}{\sqrt{3}}(2\cos x - 3\sin x) = \frac{2}{\sqrt{3}}\cos x - \frac{3}{\sqrt{3}}\sin x \]
and
\[ B = \frac{1}{\sqrt{3}}(2\cos x + 3\sin x) = \frac{2}{\sqrt{3}}\cos x + \frac{3}{\sqrt{3}}\sin x \]

### Step 2: Using Trigonometric Identities

We can express \( A \) and \( B \) in terms of angles. Let:
\[ \cos \alpha = \frac{2}{\sqrt{3}}, \quad \sin \alpha = \frac{3}{\sqrt{3}} \]
This gives:
\[ \tan \alpha = \frac{3}{2} \implies \alpha = \tan^{-1}\left(\frac{3}{2}\right) \]

Thus, we can rewrite \( A \) as:
\[ A = \cos(x - \alpha) \]

Similarly, for \( B \):
Let:
\[ \sin \beta = \frac{2}{\sqrt{3}}, \quad \cos \beta = \frac{3}{\sqrt{3}} \]
This gives:
\[ \tan \beta = \frac{2}{3} \implies \beta = \tan^{-1}\left(\frac{2}{3}\right) \]

Thus, we can rewrite \( B \) as:
\[ B = \sin(x + \beta) \]

### Step 3: Combining the Functions

Now we can express \( f(x) \) as:
\[ f(x) = \cos^{-1}(\cos(x - \alpha)) + \sin^{-1}(\sin(x + \beta)) \]

Using the properties of inverse functions, we have:
\[ f(x) = (x - \alpha) + (x + \beta) = 2x - \alpha + \beta \]

### Step 4: Differentiating \( f(x) \)

Now, we differentiate \( f(x) \):
\[ f'(x) = 2 \]

### Step 5: Finding \( g'(x) \)

Next, we need to differentiate \( g(x) = \sqrt{1 + x^2} \):
\[ g'(x) = \frac{x}{\sqrt{1 + x^2}} \]

### Step 6: Evaluating at \( x = \frac{1}{\sqrt{3}} \)

Now we evaluate \( f' \) and \( g' \) at \( x = \frac{1}{\sqrt{3}} \):
\[ f'\left(\frac{1}{\sqrt{3}}\right) = 2 \]
\[ g'\left(\frac{1}{\sqrt{3}}\right) = \frac{\frac{1}{\sqrt{3}}}{\sqrt{1 + \left(\frac{1}{\sqrt{3}}\right)^2}} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{1 + \frac{1}{3}}} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{\frac{4}{3}}} = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} \]

### Step 7: Finding the Final Derivative

Now we find the derivative of \( f \) with respect to \( g \):
\[ \frac{df}{dg} = \frac{f'(x)}{g'(x)} = \frac{2}{\frac{1}{2}} = 4 \]

### Final Answer

Thus, the derivative of \( f(x) \) with respect to \( \sqrt{1 + x^2} \) at \( x = \frac{1}{\sqrt{3}} \) is:

\[ \boxed{4} \]