How to find the derivative of this

[Wahamuka]

I’m confused with how to answer this problem, I’ve answered everything in a worksheet except this
Here is the question and what my current answer is in the pictures





P.s. the worksheet said the answer is (2x) - ((1)/(2xln2))

 

[topsquark]

I'd start by "simplifying" a bit. Raise both sides to the base 2:
\(\displaystyle 2^{g(x)} = \dfrac{2^{x^2}}{\sqrt{x}}\)

Now take your derivative implicitly.

-Dan

 

[HallsofIvy]

\(\displaystyle log_2\left(\frac{2^{x^2}}{\sqrt{x}}\right)= log_2\left(2^{x^2}\right)- log_2\left(x^{1/2}\right)= x^2- \frac{1}{2} log_2(x)\).

Of course, the derivative of \(\displaystyle x^2\) is 2x.

Now, what about the derivative of \(\displaystyle log_2(x)\)? If \(\displaystyle y= log_2(x)\) then \(\displaystyle x= 2^y\).

I presume that you know that the derivative of \(\displaystyle e^y\) is \(\displaystyle e^y\) again. And \(\displaystyle 2^y= e^{ln(2^y)}= e^{y ln(2)}\) so the derivative of \(\displaystyle 2^y\) with respect to x is \(\displaystyle ln(2)e^{y ln(2)}= ln(2)2^y\). (In general, the derivative of \(\displaystyle a^y\) is \(\displaystyle ln(a) a^y\).) So \(\displaystyle \frac{dx}{dy}= ln(2) 2^y\) and then \(\displaystyle \frac{dy}{dx}= \frac{1}{ln(2)}\frac{1}{2^y}= \frac{1}{ln(2)}\frac{1}{x}\). (And, in general, the derivative of \(\displaystyle log_a(x)\) is \(\displaystyle \frac{1}{ln(a)}\frac{1}{x}\).)