How do you find the derivative of #f(x) = 3x^2 ln 2x#?

Question

Christopher Agresta · Accepted Answer

#f'(x)=(dy)/(dx)=6xln2x+3x#

We must apply the product rule.

#d/(dx)(color(red)(u)v)=vcolor(red)((du)/(dx))+color(red)(u)(dv)/(dx)#

#d/(dx)(color(red)(3x^2)ln2x)#

#(dy)/(dx)=ln2xcolor(red)(d/(dx)(3x^2))+color(red)(3x^2)d/(dx)(ln2x)#

#(dy)/(dx)=ln2x xxcolor(red)( 6x)+color(red)(3x^2) xx 1/(2x)xx2#

#(dy)/(dx)=6xln2x+3x#

Christopher Allers · Answer

undefined To find the derivative of $ f(x) = 3x^2 \ln(2x) $, we can use the product rule and the chain rule. The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

Applying the product rule and the chain rule, we have:

$$ \begin{aligned} f'(x) &= \frac{d}{dx}(3x^2) \cdot \ln(2x) + 3x^2 \cdot \frac{d}{dx}(\ln(2x)) \
&= 6x \cdot \ln(2x) + 3x^2 \cdot \frac{1}{2x} \
&= 6x \cdot \ln(2x) + \frac{3x^2}{2x} \
&= 6x \cdot \ln(2x) + \frac{3x}{2}.
\end{aligned} $$