Answer :
To find the derivative of the function [tex]\( f(x) = (4x - 3)(4x^3 - x^2 + 1) \)[/tex], we can use the product rule. The product rule states that if you have two functions [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex], then the derivative of their product is:
[tex]\[
(uv)' = u'v + uv'
\][/tex]
Let's define:
- [tex]\( u(x) = 4x - 3 \)[/tex]
- [tex]\( v(x) = 4x^3 - x^2 + 1 \)[/tex]
Now, we need to find the derivatives of [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex]:
1. Derivative of [tex]\( u(x) \)[/tex]:
[tex]\( u(x) = 4x - 3 \)[/tex]
[tex]\( u'(x) = 4 \)[/tex]
2. Derivative of [tex]\( v(x) \)[/tex]:
[tex]\( v(x) = 4x^3 - x^2 + 1 \)[/tex]
To differentiate [tex]\( v(x) \)[/tex], we differentiate each term:
- The derivative of [tex]\( 4x^3 \)[/tex] is [tex]\( 12x^2 \)[/tex].
- The derivative of [tex]\( -x^2 \)[/tex] is [tex]\( -2x \)[/tex].
- The derivative of the constant [tex]\( 1 \)[/tex] is [tex]\( 0 \)[/tex].
So, [tex]\( v'(x) = 12x^2 - 2x \)[/tex].
Now, apply the product rule:
[tex]\[
f'(x) = u'(x)v(x) + u(x)v'(x)
\][/tex]
Substitute the values:
[tex]\[
f'(x) = 4(4x^3 - x^2 + 1) + (4x - 3)(12x^2 - 2x)
\][/tex]
Let's expand each term:
1. Expand [tex]\( 4(4x^3 - x^2 + 1) \)[/tex]:
- [tex]\( 4 \times 4x^3 = 16x^3 \)[/tex]
- [tex]\( 4 \times -x^2 = -4x^2 \)[/tex]
- [tex]\( 4 \times 1 = 4 \)[/tex]
So, the expanded form is [tex]\( 16x^3 - 4x^2 + 4 \)[/tex].
2. Expand [tex]\( (4x - 3)(12x^2 - 2x) \)[/tex]:
- [tex]\( 4x \times 12x^2 = 48x^3 \)[/tex]
- [tex]\( 4x \times -2x = -8x^2 \)[/tex]
- [tex]\( -3 \times 12x^2 = -36x^2 \)[/tex]
- [tex]\( -3 \times -2x = 6x \)[/tex]
Combine these results:
[tex]\[
48x^3 - 8x^2 - 36x^2 + 6x = 48x^3 - 44x^2 + 6x
\][/tex]
Now, combine the two expanded forms:
[tex]\[
f'(x) = (16x^3 - 4x^2 + 4) + (48x^3 - 44x^2 + 6x)
\][/tex]
[tex]\[
f'(x) = 16x^3 - 4x^2 + 4 + 48x^3 - 44x^2 + 6x
\][/tex]
Combine like terms:
- [tex]\( 16x^3 + 48x^3 = 64x^3 \)[/tex]
- [tex]\( -4x^2 - 44x^2 = -48x^2 \)[/tex]
- [tex]\( 6x \)[/tex] remains as it is.
- The constant term is [tex]\( 4 \)[/tex].
So, the derivative is:
[tex]\[
f'(x) = 64x^3 - 48x^2 + 6x + 4
\][/tex]
This is the final result.
[tex]\[
(uv)' = u'v + uv'
\][/tex]
Let's define:
- [tex]\( u(x) = 4x - 3 \)[/tex]
- [tex]\( v(x) = 4x^3 - x^2 + 1 \)[/tex]
Now, we need to find the derivatives of [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex]:
1. Derivative of [tex]\( u(x) \)[/tex]:
[tex]\( u(x) = 4x - 3 \)[/tex]
[tex]\( u'(x) = 4 \)[/tex]
2. Derivative of [tex]\( v(x) \)[/tex]:
[tex]\( v(x) = 4x^3 - x^2 + 1 \)[/tex]
To differentiate [tex]\( v(x) \)[/tex], we differentiate each term:
- The derivative of [tex]\( 4x^3 \)[/tex] is [tex]\( 12x^2 \)[/tex].
- The derivative of [tex]\( -x^2 \)[/tex] is [tex]\( -2x \)[/tex].
- The derivative of the constant [tex]\( 1 \)[/tex] is [tex]\( 0 \)[/tex].
So, [tex]\( v'(x) = 12x^2 - 2x \)[/tex].
Now, apply the product rule:
[tex]\[
f'(x) = u'(x)v(x) + u(x)v'(x)
\][/tex]
Substitute the values:
[tex]\[
f'(x) = 4(4x^3 - x^2 + 1) + (4x - 3)(12x^2 - 2x)
\][/tex]
Let's expand each term:
1. Expand [tex]\( 4(4x^3 - x^2 + 1) \)[/tex]:
- [tex]\( 4 \times 4x^3 = 16x^3 \)[/tex]
- [tex]\( 4 \times -x^2 = -4x^2 \)[/tex]
- [tex]\( 4 \times 1 = 4 \)[/tex]
So, the expanded form is [tex]\( 16x^3 - 4x^2 + 4 \)[/tex].
2. Expand [tex]\( (4x - 3)(12x^2 - 2x) \)[/tex]:
- [tex]\( 4x \times 12x^2 = 48x^3 \)[/tex]
- [tex]\( 4x \times -2x = -8x^2 \)[/tex]
- [tex]\( -3 \times 12x^2 = -36x^2 \)[/tex]
- [tex]\( -3 \times -2x = 6x \)[/tex]
Combine these results:
[tex]\[
48x^3 - 8x^2 - 36x^2 + 6x = 48x^3 - 44x^2 + 6x
\][/tex]
Now, combine the two expanded forms:
[tex]\[
f'(x) = (16x^3 - 4x^2 + 4) + (48x^3 - 44x^2 + 6x)
\][/tex]
[tex]\[
f'(x) = 16x^3 - 4x^2 + 4 + 48x^3 - 44x^2 + 6x
\][/tex]
Combine like terms:
- [tex]\( 16x^3 + 48x^3 = 64x^3 \)[/tex]
- [tex]\( -4x^2 - 44x^2 = -48x^2 \)[/tex]
- [tex]\( 6x \)[/tex] remains as it is.
- The constant term is [tex]\( 4 \)[/tex].
So, the derivative is:
[tex]\[
f'(x) = 64x^3 - 48x^2 + 6x + 4
\][/tex]
This is the final result.
