Question

find the derivative of the function.\\( f(x) = 8x^2 ln(8x) \\)\\( f(x)= 16x^2ln(8x) + \frac{8x^x}{8x} \\)\\( \times \\)\
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find the derivative of the function.\\( f(x) = ln\left( \frac{x}{x^2 + 4} \
ight) \\)\\( f(x) = \frac{1}{x} - \frac{2x}{x^2 + 4} \\)\\( \checkmark \\)\
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Explanation:

Step1: Identify product rule form

The function is $f(x) = 8x^2 \ln(8x)$, which is a product of $u(x)=8x^2$ and $v(x)=\ln(8x)$. The product rule is $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.

Step2: Compute $u'(x)$

Differentiate $u(x)=8x^2$ using power rule:
$u'(x) = 8 \cdot 2x = 16x$

Step3: Compute $v'(x)$

Differentiate $v(x)=\ln(8x)$ using chain rule:
$v'(x) = \frac{1}{8x} \cdot 8 = \frac{1}{x}$

Step4: Apply product rule

Substitute $u, u', v, v'$ into product rule:
$f'(x) = 16x \cdot \ln(8x) + 8x^2 \cdot \frac{1}{x}$

Step5: Simplify the second term

Simplify $8x^2 \cdot \frac{1}{x}$:
$8x^2 \cdot \frac{1}{x} = 8x$

Answer:

$f'(x) = 16x\ln(8x) + 8x$