Question Statement

Given the function:

$f (x) = (x^{2} + 1) ln (x^{2} + 1)$

Find the first derivative $f^{'} (x)$ and the second derivative $f^{''} (x)$ .

Background and Explanation

This problem requires applying the product rule for differentiation, which handles functions that are products of two expressions, along with the chain rule for the composite logarithmic function $ln (x^{2} + 1)$ .

Solution

Finding the First Derivative $f^{'} (x)$

We treat $f (x)$ as a product of two functions: $u = x^{2} + 1$ and $v = ln (x^{2} + 1)$ . Applying the product rule $\frac{d}{d x} (uv) = u^{'} v + u v^{'}$ :

$f^{'} (x) = \frac{d}{d x} (x^{2} + 1) \cdot ln (x^{2} + 1) + (x^{2} + 1) \cdot \frac{d}{d x} (ln (x^{2} + 1))$

Computing the individual derivatives:

$\frac{d}{d x} (x^{2} + 1) = 2 x + 0$
$\frac{d}{d x} (ln (x^{2} + 1)) = \frac{1}{x ^{2} + 1} \cdot 2 x$ (using the chain rule)

Substituting these back:

$f^{'} (x) = (2 x + 0) ln (x^{2} + 1) + (x^{2} + 1) \cdot \frac{1}{x ^{2} + 1} \cdot 2 x$

Notice that $(x^{2} + 1)$ and $\frac{1}{x ^{2} + 1}$ cancel in the second term:

$f^{'} (x) = 2 x ln (x^{2} + 1) + 2 x$

Finding the Second Derivative $f^{''} (x)$

Now we differentiate $f^{'} (x) = 2 x ln (x^{2} + 1) + 2 x$ with respect to $x$ . We apply the product rule to the first term $2 x ln (x^{2} + 1)$ and standard differentiation to the second term $2 x$ :

$f^{''} (x) = \frac{d}{d x} (2 x) \cdot ln (x^{2} + 1) + (2 x) \cdot \frac{d}{d x} (ln (x^{2} + 1)) + \frac{d}{d x} (2 x)$

Computing each component:

$\frac{d}{d x} (2 x) = 2$
$\frac{d}{d x} (ln (x^{2} + 1)) = \frac{2 x}{x ^{2} + 1}$

Substituting these values:

$f^{''} (x) = 2 ln (x^{2} + 1) + 2 x \cdot \frac{1}{x ^{2} + 1} \cdot 2 x + 2$

Simplifying the middle term ( $2 x \cdot 2 x = 4 x^{2}$ ):

$f^{''} (x) = 2 ln (x^{2} + 1) + \frac{4 x ^{2}}{x ^{2} + 1} + 2$

Rearranging the constant term to group similar expressions:

$f^{''} (x) = 2 ln (x^{2} + 1) + 2 + \frac{4 x ^{2}}{x ^{2} + 1}$

Key Formulas or Methods Used

Product Rule: $\frac{d}{d x} [u (x) v (x)] = u^{'} (x) v (x) + u (x) v^{'} (x)$
Chain Rule: $\frac{d}{d x} [ln (u (x))] = \frac{u ^{'} ( x )}{u ( x )}$
Power Rule: $\frac{d}{d x} [x^{n}] = n x^{n - 1}$

Summary of Steps

Identify the structure: Recognize $f (x)$ as a product of $(x^{2} + 1)$ and $ln (x^{2} + 1)$
First derivative: Apply product rule to find $f^{'} (x)$ , using chain rule for the logarithmic term
Simplify $f^{'} (x)$ : Cancel the $(x^{2} + 1)$ terms to get $f^{'} (x) = 2 x ln (x^{2} + 1) + 2 x$
Second derivative setup: Differentiate $f^{'} (x)$ by applying product rule to $2 x ln (x^{2} + 1)$ and power rule to $2 x$
Compute and simplify: Calculate derivatives and combine terms to obtain $f^{''} (x) = 2 ln (x^{2} + 1) + 2 + \frac{4 x ^{2}}{x ^{2} + 1}$

Question Statement

Given the function:

$f (x) = (x^{2} + 1) ln (x^{2} + 1)$

Find the first derivative $f^{'} (x)$ and the second derivative $f^{''} (x)$ .

Background and Explanation

Solution

Finding the First Derivative $f^{'} (x)$

We treat $f (x)$ as a product of two functions: $u = x^{2} + 1$ and $v = ln (x^{2} + 1)$ . Applying the product rule $\frac{d}{d x} (uv) = u^{'} v + u v^{'}$ :

$f^{'} (x) = \frac{d}{d x} (x^{2} + 1) \cdot ln (x^{2} + 1) + (x^{2} + 1) \cdot \frac{d}{d x} (ln (x^{2} + 1))$

Computing the individual derivatives:

$\frac{d}{d x} (x^{2} + 1) = 2 x + 0$
$\frac{d}{d x} (ln (x^{2} + 1)) = \frac{1}{x ^{2} + 1} \cdot 2 x$ (using the chain rule)

Substituting these back:

$f^{'} (x) = (2 x + 0) ln (x^{2} + 1) + (x^{2} + 1) \cdot \frac{1}{x ^{2} + 1} \cdot 2 x$

Notice that $(x^{2} + 1)$ and $\frac{1}{x ^{2} + 1}$ cancel in the second term:

$f^{'} (x) = 2 x ln (x^{2} + 1) + 2 x$

Finding the Second Derivative $f^{''} (x)$

$f^{''} (x) = \frac{d}{d x} (2 x) \cdot ln (x^{2} + 1) + (2 x) \cdot \frac{d}{d x} (ln (x^{2} + 1)) + \frac{d}{d x} (2 x)$

Computing each component:

$\frac{d}{d x} (2 x) = 2$
$\frac{d}{d x} (ln (x^{2} + 1)) = \frac{2 x}{x ^{2} + 1}$

Substituting these values:

$f^{''} (x) = 2 ln (x^{2} + 1) + 2 x \cdot \frac{1}{x ^{2} + 1} \cdot 2 x + 2$

Simplifying the middle term ( $2 x \cdot 2 x = 4 x^{2}$ ):

$f^{''} (x) = 2 ln (x^{2} + 1) + \frac{4 x ^{2}}{x ^{2} + 1} + 2$

Rearranging the constant term to group similar expressions:

$f^{''} (x) = 2 ln (x^{2} + 1) + 2 + \frac{4 x ^{2}}{x ^{2} + 1}$

Key Formulas or Methods Used

Product Rule: $\frac{d}{d x} [u (x) v (x)] = u^{'} (x) v (x) + u (x) v^{'} (x)$
Chain Rule: $\frac{d}{d x} [ln (u (x))] = \frac{u ^{'} ( x )}{u ( x )}$
Power Rule: $\frac{d}{d x} [x^{n}] = n x^{n - 1}$

Summary of Steps

Identify the structure: Recognize $f (x)$ as a product of $(x^{2} + 1)$ and $ln (x^{2} + 1)$
First derivative: Apply product rule to find $f^{'} (x)$ , using chain rule for the logarithmic term
Simplify $f^{'} (x)$ : Cancel the $(x^{2} + 1)$ terms to get $f^{'} (x) = 2 x ln (x^{2} + 1) + 2 x$
Second derivative setup: Differentiate $f^{'} (x)$ by applying product rule to $2 x ln (x^{2} + 1)$ and power rule to $2 x$
Compute and simplify: Calculate derivatives and combine terms to obtain $f^{''} (x) = 2 ln (x^{2} + 1) + 2 + \frac{4 x ^{2}}{x ^{2} + 1}$

2.8 Q-14

Question Statement

Background and Explanation

Solution

Finding the First Derivative $f^{'} (x)$

Finding the Second Derivative $f^{''} (x)$

Key Formulas or Methods Used

Summary of Steps

2.8 Q-14

Question Statement

Background and Explanation

Solution

Finding the First Derivative $f^{'} (x)$

Finding the Second Derivative $f^{''} (x)$

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Finding the First Derivative f′(x)

Finding the Second Derivative f′′(x)

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Finding the First Derivative f′(x)

Finding the Second Derivative f′′(x)

Key Formulas or Methods Used

Summary of Steps

Finding the First Derivative $f^{'} (x)$

Finding the Second Derivative $f^{''} (x)$

Finding the First Derivative $f^{'} (x)$

Finding the Second Derivative $f^{''} (x)$