Answer :

We want to differentiate
[tex]$$
y = 21x^4 \ln\left(5x^2 - x\right)
$$[/tex]
with respect to [tex]$x$[/tex]. Notice that this expression is a product of two functions:
[tex]$$
u(x) = 21x^4 \quad \text{and} \quad v(x) = \ln\left(5x^2 - x\right).
$$[/tex]

### Step 1. Use the Product Rule

The product rule states that if
[tex]$$
y = u(x)v(x),
$$[/tex]
then
[tex]$$
y' = u'(x)v(x) + u(x)v'(x).
$$[/tex]

### Step 2. Differentiate [tex]$u(x)=21x^4$[/tex]

Differentiate [tex]$u(x)$[/tex] using the power rule:
[tex]$$
u'(x) = 21 \cdot 4x^3 = 84x^3.
$$[/tex]

### Step 3. Differentiate [tex]$v(x) = \ln\left(5x^2 - x\right)$[/tex]

To differentiate [tex]$v(x)$[/tex], we use the chain rule. Recall that if
[tex]$$
v(x) = \ln(g(x)),
$$[/tex]
then
[tex]$$
v'(x) = \frac{g'(x)}{g(x)}.
$$[/tex]

Here, [tex]$g(x) = 5x^2 - x$[/tex]. Now, differentiate [tex]$g(x)$[/tex]:
[tex]$$
g'(x) = \frac{d}{dx}(5x^2) - \frac{d}{dx}(x) = 10x - 1.
$$[/tex]

Thus,
[tex]$$
v'(x) = \frac{10x - 1}{5x^2 - x}.
$$[/tex]

### Step 4. Combine the Results

Now substitute [tex]$u'(x)$[/tex], [tex]$v(x)$[/tex], [tex]$u(x)$[/tex], and [tex]$v'(x)$[/tex] into the product rule formula:

[tex]\[
\begin{aligned}
y' &= u'(x)v(x) + u(x)v'(x) \\
&= 84x^3 \ln\left(5x^2 - x\right) + 21x^4 \cdot \frac{10x - 1}{5x^2 - x}.
\end{aligned}
\][/tex]

Thus, the derivative of the given expression is:
[tex]$$
y' = 84x^3 \ln\left(5x^2 - x\right) + \frac{21x^4(10x - 1)}{5x^2 - x}.
$$[/tex]

This is the final answer.

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