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To find the second derivative of the function [tex]\( f(x) = x(4x^2 - 1)^5 \)[/tex], we will start by finding the first derivative and then the second derivative. Here are the steps:
### Step 1: Find the first derivative [tex]\( f'(x) \)[/tex]
First, let’s apply the product rule, which states that if we have a function [tex]\( f(x) = u(x)v(x) \)[/tex], then its derivative is [tex]\( f'(x) = u'(x)v(x) + u(x)v'(x) \)[/tex].
In this case:
- [tex]\( u(x) = x \)[/tex]
- [tex]\( v(x) = (4x^2 - 1)^5 \)[/tex]
First, we find the derivatives of [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex]:
- The derivative of [tex]\( u(x) = x \)[/tex] is [tex]\( u'(x) = 1 \)[/tex].
- To find the derivative of [tex]\( v(x) = (4x^2 - 1)^5 \)[/tex], we use the chain rule. Let [tex]\( g(x) = 4x^2 - 1 \)[/tex] and [tex]\( h(g) = g^5 \)[/tex]. Then
[tex]\[ v'(x) = \frac{d}{dx}[(4x^2 - 1)^5] = 5(4x^2 - 1)^4 \cdot \frac{d}{dx}(4x^2 - 1) \][/tex]
[tex]\[ \frac{d}{dx}(4x^2 - 1) = 8x \][/tex]
Therefore,
[tex]\[ v'(x) = 5(4x^2 - 1)^4 \cdot 8x = 40x(4x^2 - 1)^4 \][/tex]
Now, applying the product rule:
[tex]\[ f'(x) = u'(x)v(x) + u(x)v'(x) \][/tex]
[tex]\[ f'(x) = 1 \cdot (4x^2 - 1)^5 + x \cdot 40x(4x^2 - 1)^4 \][/tex]
[tex]\[ f'(x) = (4x^2 - 1)^5 + 40x^2(4x^2 - 1)^4 \][/tex]
We can combine the terms:
[tex]\[ f'(x) = 40x^2(4x^2 - 1)^4 + (4x^2 - 1)^5 \][/tex]
### Step 2: Find the second derivative [tex]\( f''(x) \)[/tex]
Now, we need to differentiate [tex]\( f'(x) \)[/tex] to find the second derivative [tex]\( f''(x) \)[/tex].
[tex]\[ f'(x) = 40x^2(4x^2 - 1)^4 + (4x^2 - 1)^5 \][/tex]
We again apply the product rule and chain rule where necessary.
For the term [tex]\( 40x^2(4x^2 - 1)^4 \)[/tex]:
[tex]\[ \frac{d}{dx}[40x^2(4x^2 - 1)^4] = 40 \cdot \frac{d}{dx}[x^2 \cdot (4x^2 - 1)^4] \][/tex]
Let [tex]\( u(x) = x^2 \)[/tex] and [tex]\( v(x) = (4x^2 - 1)^4 \)[/tex]:
[tex]\[ \frac{d}{dx}[x^2(4x^2 - 1)^4] = x^2 \cdot \frac{d}{dx}[(4x^2 - 1)^4] + (4x^2 - 1)^4 \cdot \frac{d}{dx}[x^2] \][/tex]
We already found:
[tex]\[ \frac{d}{dx}[(4x^2 - 1)^4] = 16x(4x^2-1)^3 \][/tex]
[tex]\[ \frac{d}{dx}[x^2] = 2x \][/tex]
Thus,
[tex]\[ \frac{d}{dx}[x^2(4x^2 - 1)^4] = x^2 \cdot 16x(4x^2 - 1)^3 + 2x(4x^2 - 1)^4 \][/tex]
[tex]\[ = 16x^3(4x^2 - 1)^3 + 2x(4x^2 - 1)^4 \][/tex]
Multiply by 40:
[tex]\[ 40 \left(16x^3(4x^2 - 1)^3 + 2x(4x^2 - 1)^4\right) = 640x^3(4x^2 - 1)^3 + 80x(4x^2 - 1)^4 \][/tex]
Next, for the term [tex]\( (4x^2 - 1)^5 \)[/tex]:
[tex]\[ \frac{d}{dx}[(4x^2 - 1)^5] = 20x(4x^2 - 1)^4 \][/tex]
Putting it all together:
[tex]\[ f''(x) = 640x^3(4x^2 - 1)^3 + 80x(4x^2 - 1)^4 + 20x(4x^2 - 1)^4 \][/tex]
Combine like terms:
[tex]\[ f''(x) = 640x^3(4x^2 - 1)^3 + 100x(4x^2 - 1)^4 \][/tex]
Thus, the second derivative of the function [tex]\( f(x) = x(4x^2 - 1)^5 \)[/tex] is:
[tex]\[ f''(x) = 1280x^3(4x^2 - 1)^3 + 120x(4x^2 - 1)^4 \][/tex]
### Step 1: Find the first derivative [tex]\( f'(x) \)[/tex]
First, let’s apply the product rule, which states that if we have a function [tex]\( f(x) = u(x)v(x) \)[/tex], then its derivative is [tex]\( f'(x) = u'(x)v(x) + u(x)v'(x) \)[/tex].
In this case:
- [tex]\( u(x) = x \)[/tex]
- [tex]\( v(x) = (4x^2 - 1)^5 \)[/tex]
First, we find the derivatives of [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex]:
- The derivative of [tex]\( u(x) = x \)[/tex] is [tex]\( u'(x) = 1 \)[/tex].
- To find the derivative of [tex]\( v(x) = (4x^2 - 1)^5 \)[/tex], we use the chain rule. Let [tex]\( g(x) = 4x^2 - 1 \)[/tex] and [tex]\( h(g) = g^5 \)[/tex]. Then
[tex]\[ v'(x) = \frac{d}{dx}[(4x^2 - 1)^5] = 5(4x^2 - 1)^4 \cdot \frac{d}{dx}(4x^2 - 1) \][/tex]
[tex]\[ \frac{d}{dx}(4x^2 - 1) = 8x \][/tex]
Therefore,
[tex]\[ v'(x) = 5(4x^2 - 1)^4 \cdot 8x = 40x(4x^2 - 1)^4 \][/tex]
Now, applying the product rule:
[tex]\[ f'(x) = u'(x)v(x) + u(x)v'(x) \][/tex]
[tex]\[ f'(x) = 1 \cdot (4x^2 - 1)^5 + x \cdot 40x(4x^2 - 1)^4 \][/tex]
[tex]\[ f'(x) = (4x^2 - 1)^5 + 40x^2(4x^2 - 1)^4 \][/tex]
We can combine the terms:
[tex]\[ f'(x) = 40x^2(4x^2 - 1)^4 + (4x^2 - 1)^5 \][/tex]
### Step 2: Find the second derivative [tex]\( f''(x) \)[/tex]
Now, we need to differentiate [tex]\( f'(x) \)[/tex] to find the second derivative [tex]\( f''(x) \)[/tex].
[tex]\[ f'(x) = 40x^2(4x^2 - 1)^4 + (4x^2 - 1)^5 \][/tex]
We again apply the product rule and chain rule where necessary.
For the term [tex]\( 40x^2(4x^2 - 1)^4 \)[/tex]:
[tex]\[ \frac{d}{dx}[40x^2(4x^2 - 1)^4] = 40 \cdot \frac{d}{dx}[x^2 \cdot (4x^2 - 1)^4] \][/tex]
Let [tex]\( u(x) = x^2 \)[/tex] and [tex]\( v(x) = (4x^2 - 1)^4 \)[/tex]:
[tex]\[ \frac{d}{dx}[x^2(4x^2 - 1)^4] = x^2 \cdot \frac{d}{dx}[(4x^2 - 1)^4] + (4x^2 - 1)^4 \cdot \frac{d}{dx}[x^2] \][/tex]
We already found:
[tex]\[ \frac{d}{dx}[(4x^2 - 1)^4] = 16x(4x^2-1)^3 \][/tex]
[tex]\[ \frac{d}{dx}[x^2] = 2x \][/tex]
Thus,
[tex]\[ \frac{d}{dx}[x^2(4x^2 - 1)^4] = x^2 \cdot 16x(4x^2 - 1)^3 + 2x(4x^2 - 1)^4 \][/tex]
[tex]\[ = 16x^3(4x^2 - 1)^3 + 2x(4x^2 - 1)^4 \][/tex]
Multiply by 40:
[tex]\[ 40 \left(16x^3(4x^2 - 1)^3 + 2x(4x^2 - 1)^4\right) = 640x^3(4x^2 - 1)^3 + 80x(4x^2 - 1)^4 \][/tex]
Next, for the term [tex]\( (4x^2 - 1)^5 \)[/tex]:
[tex]\[ \frac{d}{dx}[(4x^2 - 1)^5] = 20x(4x^2 - 1)^4 \][/tex]
Putting it all together:
[tex]\[ f''(x) = 640x^3(4x^2 - 1)^3 + 80x(4x^2 - 1)^4 + 20x(4x^2 - 1)^4 \][/tex]
Combine like terms:
[tex]\[ f''(x) = 640x^3(4x^2 - 1)^3 + 100x(4x^2 - 1)^4 \][/tex]
Thus, the second derivative of the function [tex]\( f(x) = x(4x^2 - 1)^5 \)[/tex] is:
[tex]\[ f''(x) = 1280x^3(4x^2 - 1)^3 + 120x(4x^2 - 1)^4 \][/tex]
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