Question Statement

Find the first and second derivatives of the function:

$f (x) = x^{2} (3 x - 4)^{3}$

Background and Explanation

This problem requires applying the product rule for differentiation, which states that $\frac{d}{d x} [u (x) v (x)] = u^{'} (x) v (x) + u (x) v^{'} (x)$ . Since $(3 x - 4)^{3}$ is a composite function, the chain rule is also necessary to handle the inner function $(3 x - 4)$ .

Solution

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

We apply the product rule to $f (x) = x^{2} (3 x - 4)^{3}$ , treating $u = x^{2}$ and $v = (3 x - 4)^{3}$ :

$f^{'} (x) = \frac{d}{d x} (x^{2}) \cdot (3 x - 4)^{3} + x^{2} \cdot \frac{d}{d x} (3 x - 4)^{3} = 2 x (3 x - 4)^{3} + x^{2} \cdot 3 (3 x - 4)^{2} \cdot \frac{d}{d x} (3 x - 4) = 2 x (3 x - 4)^{3} + 3 x^{2} (3 x - 4)^{2} \cdot 3 = 2 x (3 x - 4)^{3} + 9 x^{2} (3 x - 4)^{2}$

Factor out the common term $(3 x - 4)^{2}$ to simplify:

$f^{'} (x) = (3 x - 4)^{2} [2 x (3 x - 4) + 9 x^{2}] = (3 x - 4)^{2} (6 x^{2} - 8 x + 9 x^{2}) = (3 x - 4)^{2} (15 x^{2} - 8 x)$

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

Differentiate $f^{'} (x) = (3 x - 4)^{2} (15 x^{2} - 8 x)$ using the product rule again:

$f^{''} (x) = \frac{d}{d x} (3 x - 4)^{2} \cdot (15 x^{2} - 8 x) + (3 x - 4)^{2} \cdot \frac{d}{d x} (15 x^{2} - 8 x) = 2 (3 x - 4) \cdot 3 \cdot (15 x^{2} - 8 x) + (3 x - 4)^{2} (30 x - 8) = 6 (3 x - 4) (15 x^{2} - 8 x) + (3 x - 4)^{2} (30 x - 8)$

Factor out $(3 x - 4)$ and expand the remaining terms:

$f^{''} (x) = (3 x - 4) [6 (15 x^{2} - 8 x) + (3 x - 4) (30 x - 8)] = (3 x - 4) [90 x^{2} - 48 x + 90 x^{2} - 24 x - 120 x + 32] = (3 x - 4) (180 x^{2} - 192 x + 32)$

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: $\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)$
Power Rule: $\frac{d}{d x} [x^{n}] = n x^{n - 1}$

Summary of Steps

First Derivative:
- Identify $u = x^{2}$ and $v = (3 x - 4)^{3}$
- Apply product rule: $u^{'} v + u v^{'}$
- Apply chain rule to find $v^{'} = 3 (3 x - 4)^{2} \cdot 3 = 9 (3 x - 4)^{2}$
- Simplify to $2 x (3 x - 4)^{3} + 9 x^{2} (3 x - 4)^{2}$
- Factor out $(3 x - 4)^{2}$ and combine like terms inside brackets to get $(3 x - 4)^{2} (15 x^{2} - 8 x)$
Second Derivative:
- Treat $f^{'} (x)$ as product of $(3 x - 4)^{2}$ and $(15 x^{2} - 8 x)$
- Apply product rule again
- Differentiate $(3 x - 4)^{2}$ using chain rule to get $6 (3 x - 4)$
- Differentiate $(15 x^{2} - 8 x)$ to get $(30 x - 8)$
- Factor out $(3 x - 4)$ and expand the remaining expression
- Combine like terms: $90 x^{2} + 90 x^{2} = 180 x^{2}$ , $- 48 x - 24 x - 120 x = - 192 x$ , and constant $32$

Finding the First Derivative

f^{'} (x)

We apply the product rule to

f (x) = x^{2} (3 x - 4)^{3}

, treating

u = x^{2}

and

v = (3 x - 4)^{3}

f^{'} (x) = \frac{d}{d x} (x^{2}) \cdot (3 x - 4)^{3} + x^{2} \cdot \frac{d}{d x} (3 x - 4)^{3} = 2 x (3 x - 4)^{3} + x^{2} \cdot 3 (3 x - 4)^{2} \cdot \frac{d}{d x} (3 x - 4) = 2 x (3 x - 4)^{3} + 3 x^{2} (3 x - 4)^{2} \cdot 3 = 2 x (3 x - 4)^{3} + 9 x^{2} (3 x - 4)^{2}

Factor out the common term

(3 x - 4)^{2}

to simplify:

f^{'} (x) = (3 x - 4)^{2} [2 x (3 x - 4) + 9 x^{2}] = (3 x - 4)^{2} (6 x^{2} - 8 x + 9 x^{2}) = (3 x - 4)^{2} (15 x^{2} - 8 x)

Finding the Second Derivative

f^{''} (x)

Differentiate

f^{'} (x) = (3 x - 4)^{2} (15 x^{2} - 8 x)

using the product rule again:

f^{''} (x) = \frac{d}{d x} (3 x - 4)^{2} \cdot (15 x^{2} - 8 x) + (3 x - 4)^{2} \cdot \frac{d}{d x} (15 x^{2} - 8 x) = 2 (3 x - 4) \cdot 3 \cdot (15 x^{2} - 8 x) + (3 x - 4)^{2} (30 x - 8) = 6 (3 x - 4) (15 x^{2} - 8 x) + (3 x - 4)^{2} (30 x - 8)

Factor out

(3 x - 4)

and expand the remaining terms:

f^{''} (x) = (3 x - 4) [6 (15 x^{2} - 8 x) + (3 x - 4) (30 x - 8)] = (3 x - 4) [90 x^{2} - 48 x + 90 x^{2} - 24 x - 120 x + 32] = (3 x - 4) (180 x^{2} - 192 x + 32)

Summary of Steps

First Derivative:

Identify $u = x^{2}$ and $v = (3 x - 4)^{3}$
Apply product rule: $u^{'} v + u v^{'}$
Apply chain rule to find $v^{'} = 3 (3 x - 4)^{2} \cdot 3 = 9 (3 x - 4)^{2}$
Simplify to $2 x (3 x - 4)^{3} + 9 x^{2} (3 x - 4)^{2}$
Factor out $(3 x - 4)^{2}$ and combine like terms inside brackets to get $(3 x - 4)^{2} (15 x^{2} - 8 x)$

Second Derivative:

Treat $f^{'} (x)$ as product of $(3 x - 4)^{2}$ and $(15 x^{2} - 8 x)$
Apply product rule again
Differentiate $(3 x - 4)^{2}$ using chain rule to get $6 (3 x - 4)$
Differentiate $(15 x^{2} - 8 x)$ to get $(30 x - 8)$
Factor out $(3 x - 4)$ and expand the remaining expression
Combine like terms: $90 x^{2} + 90 x^{2} = 180 x^{2}$ , $- 48 x - 24 x - 120 x = - 192 x$ , and constant $32$

2.8 Q-7

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-7

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 $f^{''} (x)$