Let `f(x)=|{:(cosx,x,1),(2 sinx,x^2,2x),(tanx,x,1):}|` Find `underset(xto0)"Lt"underset(x)(f'(x))`

Answer

To solve the problem, we need to find the limit of the derivative of the determinant function \( f(x) \) as \( x \) approaches 0. The function is defined as:

\[ f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x^2 & 2x \\ \tan x & x & 1 \end{array} \right| \]

### Step 1: Calculate the Determinant \( f(x) \)

We will calculate the determinant using the formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

\[ \begin{array}{ccc} a = \cos x & b = x & c = 1 \\ d = 2 \sin x & e = x^2 & f = 2x \\ g = \tan x & h = x & i = 1 \end{array} \]

The determinant can be computed as follows:

\[ f(x) = \cos x \left( x^2 \cdot 1 - 2x \cdot x \right) - x \left( 2 \sin x \cdot 1 - 2x \cdot \tan x \right) + 1 \left( 2 \sin x \cdot x - x^2 \cdot \tan x \right) \]

This simplifies to:

\[ f(x) = \cos x (x^2 - 2x^2) - x (2 \sin x - 2x \tan x) + (2x \sin x - x^2 \tan x) \]

\[ = -\cos x x^2 - x (2 \sin x - 2x \tan x) + (2x \sin x - x^2 \tan x) \]

### Step 2: Simplify \( f(x) \)

Now we can simplify \( f(x) \):

\[ f(x) = -\cos x x^2 - 2x \sin x + 2x^2 \tan x + 2x \sin x - x^2 \tan x \]

Combining like terms gives:

\[ f(x) = -\cos x x^2 + x^2 \tan x \]

### Step 3: Differentiate \( f(x) \)

Next, we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}(-\cos x x^2 + x^2 \tan x) \]

Using the product rule for differentiation:

\[ f'(x) = -\left( (-\sin x) x^2 + \cos x \cdot 2x \right) + \left( \tan x \cdot 2x + x^2 \sec^2 x \right) \]

This simplifies to:

\[ f'(x) = \sin x x^2 - 2x \cos x + 2x \tan x + x^2 \sec^2 x \]

### Step 4: Find the Limit

Now we need to find:

\[ \lim_{x \to 0} \frac{f'(x)}{x} \]

Substituting \( f'(x) \):

\[ \lim_{x \to 0} \frac{\sin x x^2 - 2x \cos x + 2x \tan x + x^2 \sec^2 x}{x} \]

This simplifies to:

\[ \lim_{x \to 0} \left( \sin x x - 2 \cos x + 2 \tan x + x \sec^2 x \right) \]

### Step 5: Evaluate the Limit

Now we evaluate the limit as \( x \to 0 \):

- \( \sin(0) = 0 \)
- \( \cos(0) = 1 \)
- \( \tan(0) = 0 \)
- \( \sec^2(0) = 1 \)

Thus, we have:

\[ \lim_{x \to 0} \left( 0 - 2 + 0 + 0 \right) = -2 \]

### Final Answer

Therefore, the limit is:

\[ \lim_{x \to 0} \frac{f'(x)}{x} = -2 \]