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 will apply the chain rule and product rule appropriately. Let's break this down step-by-step.
### Step 1: Differentiate [tex]\(\frac{6}{\sqrt{2-3x}}\)[/tex]
Rewrite the term [tex]\(\frac{6}{\sqrt{2-3x}}\)[/tex] with a negative exponent:
[tex]\[
\frac{6}{\sqrt{2-3x}} = 6(2 - 3x)^{-\frac{1}{2}}
\][/tex]
To differentiate [tex]\(6(2 - 3x)^{-\frac{1}{2}}\)[/tex], we use the chain rule:
[tex]\[
\frac{d}{dx}\left[ 6(2 - 3x)^{-\frac{1}{2}} \right] = 6 \cdot -\frac{1}{2} (2 - 3x)^{-\frac{3}{2}} \cdot (-3)
\][/tex]
Simplify:
[tex]\[
6 \cdot -\frac{1}{2} \cdot -3 \cdot (2 - 3x)^{-\frac{3}{2}} = 9(2 - 3x)^{-\frac{3}{2}}
\][/tex]
### Step 2: Differentiate [tex]\(x^2(3x^2 - x + 7)^3\)[/tex]
We need to apply the product rule here. Let [tex]\( u = x^2 \)[/tex] and [tex]\( v = (3x^2 - x + 7)^3 \)[/tex].
First, compute the derivatives:
[tex]\[
\frac{du}{dx} = \frac{d}{dx} [x^2] = 2x
\][/tex]
Now for [tex]\(v\)[/tex], we use the chain rule:
[tex]\[
\frac{dv}{dx} = \frac{d}{dx} \left[ (3x^2 - x + 7)^3 \right] = 3 \left(3x^2 - x + 7\right)^2 \cdot \frac{d}{dx} [3x^2 - x + 7]
\][/tex]
Now, differentiate the inner function [tex]\(3x^2 - x + 7\)[/tex]:
[tex]\[
\frac{d}{dx} [3x^2 - x + 7] = 6x - 1
\][/tex]
Thus:
[tex]\[
\frac{dv}{dx} = 3 \left(3x^2 - x + 7\right)^2 \cdot (6x - 1)
\][/tex]
Now apply the product rule:
[tex]\[
\frac{d}{dx} \left[x^2 (3x^2 - x + 7)^3\right] = x^2 \cdot \frac{d}{dx}[(3x^2 - x + 7)^3] + (3x^2 - x + 7)^3 \cdot \frac{d}{dx}[x^2]
\][/tex]
Substitute the derivatives:
[tex]\[
= x^2 \cdot \left(3(3x^2 - x + 7)^2 \cdot (6x - 1)\right) + (3x^2 - x + 7)^3 \cdot 2x
\][/tex]
### Step 3: Combine the Results
Add the derivatives from Steps 1 and 2 to get the final result:
[tex]\[
\frac{d}{dx}\left[\frac{6}{\sqrt{2-3x}} + x^2 (3x^2 - x + 7)^3 \right] = 9(2 - 3x)^{-\frac{3}{2}} + x^2 \cdot (18x - 3) \cdot (3x^2 - x + 7)^2 + 2x \cdot (3x^2 - x + 7)^3
\][/tex]
Therefore, the derivative is:
[tex]\[
D_x\left[\frac{6}{\sqrt{2-3x}} + x^2 \left(3x^2 - x + 7\right)^3\right] = x^2 (18x - 3) (3x^2 - x + 7)^2 + 2x (3x^2 - x + 7)^3 + \frac{9}{(2 - 3x)^{\frac{3}{2}}}
\][/tex]
