Answer :
To determine the derivative [tex]\(D_x\left[\frac{6}{\sqrt{2-3x}} + x^2\left(3x^2 - x + 7\right)^3 \right]\)[/tex], we need to use the rules of differentiation. Let's break down the given function into two parts and differentiate each part separately:
[tex]\[ f(x) = \frac{6}{\sqrt{2-3x}} + x^2 \left(3x^2 - x + 7\right)^3 \][/tex]
### Part 1: Differentiating [tex]\(\frac{6}{\sqrt{2-3x}}\)[/tex]
First, we rewrite [tex]\(\frac{6}{\sqrt{2-3x}}\)[/tex] in a more convenient form for differentiation:
[tex]\[ \frac{6}{\sqrt{2-3x}} = 6(2 - 3x)^{-\frac{1}{2}} \][/tex]
Now, we apply the chain rule to differentiate [tex]\(6(2 - 3x)^{-\frac{1}{2}}\)[/tex]:
1. Differentiate the outer function with respect to the inner function:
[tex]\[ 6 \cdot \left(-\frac{1}{2}\right) (2 - 3x)^{-\frac{3}{2}} \][/tex]
2. Differentiate the inner function [tex]\(2 - 3x\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[ -3 \][/tex]
Multiplying these together, we get:
[tex]\[ 6 \cdot \left(-\frac{1}{2}\right) (2 - 3x)^{-\frac{3}{2}} \cdot (-3) = \frac{9}{(2-3x)^{\frac{3}{2}}} \][/tex]
### Part 2: Differentiating [tex]\(x^2 \left(3x^2 - x + 7\right)^3\)[/tex]
We need to use the product rule for this part. Let:
[tex]\[ u = x^2 \][/tex]
[tex]\[ v = \left(3x^2 - x + 7\right)^3 \][/tex]
According to the product rule, the derivative of [tex]\(uv\)[/tex] is:
[tex]\[ \frac{d}{dx}[uv] = u'v + uv' \][/tex]
1. Differentiate [tex]\(u = x^2\)[/tex]:
[tex]\[ u' = 2x \][/tex]
2. Differentiate [tex]\(v = \left(3x^2 - x + 7\right)^3\)[/tex] using the chain rule:
- Let [tex]\(g(x) = 3x^2 - x + 7\)[/tex], then [tex]\(v = g(x)^3\)[/tex]
- Differentiate [tex]\(g(x)^3\)[/tex]:
[tex]\[ g'(x) = 6x - 1 \][/tex]
[tex]\[ v' = 3 \left(3x^2 - x + 7\right)^2 \cdot \left(6x - 1\right) \][/tex]
Now, applying the product rule:
[tex]\[ u'v + uv' = (2x)(3x^2 - x + 7)^3 + x^2 \cdot 3 \left(3x^2 - x + 7\right)^2 \cdot (6x - 1) \][/tex]
Simplifying this gives:
[tex]\[ 2x \left(3x^2 - x + 7\right)^3 + x^2 \cdot 3 \left(3x^2 - x + 7\right)^2 \cdot (6x - 1) \][/tex]
### Combining Both Parts
Combine the derivatives of both parts to get the final result:
[tex]\[ \frac{d}{dx} \left( \frac{6}{\sqrt{2-3x}} + x^2 \left(3x^2 - x + 7\right)^3 \right) = \frac{9}{(2-3x)^{\frac{3}{2}}} + 2x \left(3x^2 - x + 7\right)^3 + x^2 \cdot 3 \left(3x^2 - x + 7\right)^2 \cdot (6x - 1) \][/tex]
Simplifying the entire expression, we get:
[tex]\[ \frac{9}{(2-3x)^{\frac{3}{2}}} + 2x \left(3x^2 - x + 7\right)^3 + x^2(18x - 3) \left(3x^2 - x + 7\right)^2 \][/tex]
Combining like terms:
[tex]\[ \boxed{x^2(18x - 3) \left(3x^2 - x + 7\right)^2 + 2x \left(3x^2 - x + 7\right)^3 + \frac{9}{(2-3x)^{\frac{3}{2}}}} \][/tex]
This is the final result for the derivative.
[tex]\[ f(x) = \frac{6}{\sqrt{2-3x}} + x^2 \left(3x^2 - x + 7\right)^3 \][/tex]
### Part 1: Differentiating [tex]\(\frac{6}{\sqrt{2-3x}}\)[/tex]
First, we rewrite [tex]\(\frac{6}{\sqrt{2-3x}}\)[/tex] in a more convenient form for differentiation:
[tex]\[ \frac{6}{\sqrt{2-3x}} = 6(2 - 3x)^{-\frac{1}{2}} \][/tex]
Now, we apply the chain rule to differentiate [tex]\(6(2 - 3x)^{-\frac{1}{2}}\)[/tex]:
1. Differentiate the outer function with respect to the inner function:
[tex]\[ 6 \cdot \left(-\frac{1}{2}\right) (2 - 3x)^{-\frac{3}{2}} \][/tex]
2. Differentiate the inner function [tex]\(2 - 3x\)[/tex] with respect to [tex]\(x\)[/tex]:
[tex]\[ -3 \][/tex]
Multiplying these together, we get:
[tex]\[ 6 \cdot \left(-\frac{1}{2}\right) (2 - 3x)^{-\frac{3}{2}} \cdot (-3) = \frac{9}{(2-3x)^{\frac{3}{2}}} \][/tex]
### Part 2: Differentiating [tex]\(x^2 \left(3x^2 - x + 7\right)^3\)[/tex]
We need to use the product rule for this part. Let:
[tex]\[ u = x^2 \][/tex]
[tex]\[ v = \left(3x^2 - x + 7\right)^3 \][/tex]
According to the product rule, the derivative of [tex]\(uv\)[/tex] is:
[tex]\[ \frac{d}{dx}[uv] = u'v + uv' \][/tex]
1. Differentiate [tex]\(u = x^2\)[/tex]:
[tex]\[ u' = 2x \][/tex]
2. Differentiate [tex]\(v = \left(3x^2 - x + 7\right)^3\)[/tex] using the chain rule:
- Let [tex]\(g(x) = 3x^2 - x + 7\)[/tex], then [tex]\(v = g(x)^3\)[/tex]
- Differentiate [tex]\(g(x)^3\)[/tex]:
[tex]\[ g'(x) = 6x - 1 \][/tex]
[tex]\[ v' = 3 \left(3x^2 - x + 7\right)^2 \cdot \left(6x - 1\right) \][/tex]
Now, applying the product rule:
[tex]\[ u'v + uv' = (2x)(3x^2 - x + 7)^3 + x^2 \cdot 3 \left(3x^2 - x + 7\right)^2 \cdot (6x - 1) \][/tex]
Simplifying this gives:
[tex]\[ 2x \left(3x^2 - x + 7\right)^3 + x^2 \cdot 3 \left(3x^2 - x + 7\right)^2 \cdot (6x - 1) \][/tex]
### Combining Both Parts
Combine the derivatives of both parts to get the final result:
[tex]\[ \frac{d}{dx} \left( \frac{6}{\sqrt{2-3x}} + x^2 \left(3x^2 - x + 7\right)^3 \right) = \frac{9}{(2-3x)^{\frac{3}{2}}} + 2x \left(3x^2 - x + 7\right)^3 + x^2 \cdot 3 \left(3x^2 - x + 7\right)^2 \cdot (6x - 1) \][/tex]
Simplifying the entire expression, we get:
[tex]\[ \frac{9}{(2-3x)^{\frac{3}{2}}} + 2x \left(3x^2 - x + 7\right)^3 + x^2(18x - 3) \left(3x^2 - x + 7\right)^2 \][/tex]
Combining like terms:
[tex]\[ \boxed{x^2(18x - 3) \left(3x^2 - x + 7\right)^2 + 2x \left(3x^2 - x + 7\right)^3 + \frac{9}{(2-3x)^{\frac{3}{2}}}} \][/tex]
This is the final result for the derivative.
