Question

Find the derivative of y=5^xlog2(3x^2)
The log2, the 2 is like an exponent but at the bottom of the base (log). I'm not sure what the word for that is.

Answer 1

The word you're looking for is subscript. Exponents are superscripts.

Now, you have the product of two functions, so you'll be using the power rule here:
\[\begin{align*}
y&=\color{red}{5^x}\color{blue}{\log_2\left(3x^2\right)}\\
\frac{dy}{dx}&=\color{red}{\frac{d}{dx}\left[5^x\right]}\cdot\log_2\left(3x^2\right) + 5^x\cdot\color{blue}{\frac{d}{dx}\left[\log_2\left(3x^2\right)\right]}\end{align*}\]
I never remember the rules for the derivatives of these types of functions, so I'll just derive them.

Red part: Let \(s=5^x\). Then,
\[\begin{align*}\ln s&=\ln5^x\\
\ln s&=x\ln5\\
\frac{s'}{s}&=\ln5&&\text{(differentiating both sides with respect to $x$)}\\
s'&=s\ln5\\
s'&=5^x\ln5\end{align*}\]
Blue part: Let \(t=\log_2\left(3x^2\right)\). Then,
\[\begin{align*}
\large 2^t&=\large 2^{\log_2\left(3x^2\right)}\\
2^t&=3x^2\\
\large t'~2^t\ln2&=6x&&\text{using the rule from before}\\
& &&\text{(again, implicit differentiation)}\\
t'&=\frac{1}{2^t\ln2}6x\\
t'&=\frac{1}{3x^2 \ln2}6x\\
t'&=\frac{2}{x \ln2}
\end{align*}\]
So,
\[\begin{align*}
y&=\color{red}{5^x}\color{blue}{\log_2\left(3x^2\right)}\\
\frac{dy}{dx}&=\color{red}{\frac{d}{dx}\left[5^x\right]}\cdot\log_2\left(3x^2\right) + 5^x\cdot\color{blue}{\frac{d}{dx}\left[\log_2\left(3x^2\right)\right]}\\
\frac{dy}{dx}&=s'\cdot\log_2\left(3x^2\right) + 5^x\cdot t'\\
\frac{dy}{dx}&=\left(5^x\ln5\right)\cdot\log_2\left(3x^2\right) + 5^x\cdot \left(\frac{2}{x \ln2}\right)
\end{align*}\]
There a few ways to rewrite this. You could factor out the \(5^x\), for one thing. I'll leave that to you.