Question Statement

Find the first and second derivatives of the function:

$f (x) = (x^{2} + 5 x - 1)^{4}$

Background and Explanation

This problem involves differentiating a composite function using the chain rule for the first derivative, followed by the product rule combined with the chain rule for the second derivative. You should be comfortable identifying inner and outer functions and applying multiple differentiation rules sequentially.

Solution

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

We begin by recognizing that $f (x)$ is a composite function where the outer function is $u^{4}$ and the inner function is $u = x^{2} + 5 x - 1$ . Applying the chain rule $\frac{d}{d x} [u^{n}] = n u^{n - 1} \cdot \frac{d u}{d x}$ :

$f^{'} (x) = 4 (x^{2} + 5 x - 1)^{4 - 1} \cdot \frac{d}{d x} (x^{2} + 5 x - 1)$

Differentiating the inner function: $\frac{d}{d x} (x^{2} + 5 x - 1) = 2 x + 5 - 0 = 2 x + 5$

Substituting back: $f^{'} (x) = 4 (x^{2} + 5 x - 1)^{3} \cdot (2 x + 5)$

Therefore: $f^{'} (x) = 4 (2 x + 5) (x^{2} + 5 x - 1)^{3}$

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

To find $f^{''} (x)$ , we differentiate $f^{'} (x)$ with respect to $x$ . Since $f^{'} (x) = 4 (2 x + 5) (x^{2} + 5 x - 1)^{3}$ is a product of two functions, we apply the product rule $\frac{d}{d x} [uv] = u^{'} v + u v^{'}$ , where:

$u = 2 x + 5$ so $u^{'} = 2$
$v = (x^{2} + 5 x - 1)^{3}$ so $v^{'} = 3 (x^{2} + 5 x - 1)^{2} \cdot (2 x + 5)$ (using chain rule)

Applying the product rule: $f^{''} (x) = 4 [\frac{d}{d x} (2 x + 5) \cdot (x^{2} + 5 x - 1)^{3} + (2 x + 5) \cdot \frac{d}{d x} (x^{2} + 5 x - 1)^{3}]$

$= 4 [(2 + 0) (x^{2} + 5 x - 1)^{3} + (2 x + 5) \cdot 3 (x^{2} + 5 x - 1)^{2} \cdot \frac{d}{d x} (x^{2} + 5 x - 1)]$

$= 4 [2 (x^{2} + 5 x - 1)^{3} + (2 x + 5) \cdot 3 (x^{2} + 5 x - 1)^{2} (2 x + 5 - 0)]$

Factor out the common term $(x^{2} + 5 x - 1)^{2}$ : $= 4 (x^{2} + 5 x - 1)^{2} [2 (x^{2} + 5 x - 1) + 3 (2 x + 5)^{2}]$

Now expand the terms inside the brackets. First, expand $(2 x + 5)^{2}$ : $(2 x + 5)^{2} = 4 x^{2} + 20 x + 25$

So: $= 4 (x^{2} + 5 x - 1)^{2} [2 x^{2} + 10 x - 2 + 3 (4 x^{2} + 20 x + 25)]$

$= 4 (x^{2} + 5 x - 1)^{2} [2 x^{2} + 10 x - 2 + 12 x^{2} + 60 x + 75]$

Combine like terms inside the brackets: $= 4 (x^{2} + 5 x - 1)^{2} [14 x^{2} + 70 x + 73]$

Therefore, the second derivative is: $f^{''} (x) = 4 (x^{2} + 5 x - 1)^{2} (14 x^{2} + 70 x + 73)$

Key Formulas or Methods Used

Chain Rule: $\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)$ or $\frac{d}{d x} [u^{n}] = n u^{n - 1} \cdot \frac{d u}{d x}$
Product Rule: $\frac{d}{d x} [u \cdot v] = u^{'} v + u v^{'}$
Power Rule: $\frac{d}{d x} [x^{n}] = n x^{n - 1}$
Algebraic Expansion: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$

Summary of Steps

Apply the chain rule to $f (x) = (x^{2} + 5 x - 1)^{4}$ to find $f^{'} (x) = 4 (2 x + 5) (x^{2} + 5 x - 1)^{3}$
Recognize $f^{'} (x)$ as a product of $(2 x + 5)$ and $(x^{2} + 5 x - 1)^{3}$ for the second derivative
Apply the product rule, using the chain rule again to differentiate the second factor
Factor out common terms $(x^{2} + 5 x - 1)^{2}$ to simplify the expression
Expand $(2 x + 5)^{2}$ and distribute the 3 inside the brackets
Combine like terms to obtain the final simplified form of $f^{''} (x)$