Let `f(x)=tan^(-1)((x^(3)-1)/(x^(2)+x))`, then the value of `17f'(2)` is equal to

Answer

To solve the problem, we need to find the value of \( 17f'(2) \) where \( f(x) = \tan^{-1}\left(\frac{x^3 - 1}{x^2 + x}\right) \).

### Step 1: Differentiate \( f(x) \)

We will use the chain rule and the derivative of the inverse tangent function. The derivative of \( \tan^{-1}(u) \) is given by:

\[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]

Here, \( u = \frac{x^3 - 1}{x^2 + x} \). We need to find \( \frac{du}{dx} \).

### Step 2: Find \( \frac{du}{dx} \)

Using the quotient rule:

\[ \frac{du}{dx} = \frac{(x^2 + x)(3x^2) - (x^3 - 1)(2x + 1)}{(x^2 + x)^2} \]

### Step 3: Simplify \( \frac{du}{dx} \)

Calculating the numerator:

1. \( (x^2 + x)(3x^2) = 3x^4 + 3x^3 \)
2. \( (x^3 - 1)(2x + 1) = 2x^4 + x^3 - 2x - 1 \)

Now, combine these:

\[ \text{Numerator} = 3x^4 + 3x^3 - (2x^4 + x^3 - 2x - 1) = 3x^4 + 3x^3 - 2x^4 - x^3 + 2x + 1 = x^4 + 2x^3 + 2x + 1 \]

Thus,

\[ \frac{du}{dx} = \frac{x^4 + 2x^3 + 2x + 1}{(x^2 + x)^2} \]

### Step 4: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative of \( f(x) \)

Now we can find \( f'(x) \):

\[ f'(x) = \frac{1}{1 + \left(\frac{x^3 - 1}{x^2 + x}\right)^2} \cdot \frac{x^4 + 2x^3 + 2x + 1}{(x^2 + x)^2} \]

### Step 5: Evaluate \( f'(2) \)

First, we need to find \( u \) at \( x = 2 \):

\[ u = \frac{2^3 - 1}{2^2 + 2} = \frac{8 - 1}{4 + 2} = \frac{7}{6} \]

Now calculate \( 1 + u^2 \):

\[ 1 + u^2 = 1 + \left(\frac{7}{6}\right)^2 = 1 + \frac{49}{36} = \frac{36 + 49}{36} = \frac{85}{36} \]

Now substitute \( x = 2 \) into \( \frac{du}{dx} \):

\[ \frac{du}{dx} \text{ at } x = 2 = \frac{2^4 + 2 \cdot 2^3 + 2 \cdot 2 + 1}{(2^2 + 2)^2} = \frac{16 + 16 + 4 + 1}{36} = \frac{37}{36} \]

Now substitute into \( f'(2) \):

\[ f'(2) = \frac{1}{\frac{85}{36}} \cdot \frac{37}{36} = \frac{36}{85} \cdot \frac{37}{36} = \frac{37}{85} \]

### Step 6: Calculate \( 17f'(2) \)

Finally, we find:

\[ 17f'(2) = 17 \cdot \frac{37}{85} = \frac{629}{85} \]

### Final Answer

The value of \( 17f'(2) \) is \( \frac{629}{85} \).