# Need help on this question. thanks guys

• Jun 27th 2014, 02:32 AM
pumbaa213
Need help on this question. thanks guys
(i) Find the numerical value of the derivative of 3^x when x=2.
The answer given is 9.89

How do i solve this?

My attempt:
Firstly, i equate y=3^x
then i tried using ln to solve but i got stuck somewhere..
lny=xln3
...

• Jun 27th 2014, 04:31 AM
pumbaa213
Re: Need help on this question. thanks guys
its okay guys. I managed to solve it (finally!)

Let y=3^x
y=3^x
lny=ln3^x (ln both sides)
lny=xln3 (due to the law of log: lnx^a = alnx)
y= e^xln3 (change to index form)
dy/dx= e^xln3(ln3) (differentiate y)
dy/dx = 9.89 (3 s.f.)
• Jun 27th 2014, 06:55 AM
HallsofIvy
Re: Need help on this question. thanks guys
Another way to do this: $y= 3^x$ so ln(y)= x ln(3). Differentiate both sides with respect to x:
$\frac{1}{y}\frac{dy}{dx}= ln(3)$ so $\frac{dy}{dx}= y ln(3)= 3^x ln(3)$.

Going back to the definition of derivative, the difference quotient would be $\frac{3^{x+h}- 3^x}{h}= \frac{3^x3^h- 3^x}{h}= 3^x\frac{3^h- 1}{h}$. So the derivative is given by $\lim_{h\to 0} 3^x \frac{3^h- 1}{h}= 3^x\left(\lim_{h\to 0}\frac{3^h- 1}{h}\right)$

Notice that the first part of that, $3^x$ does not have an "h" while the limit does not involve "x". That means that the derivative of $3^x$ is just $3^x$ times a constant. (Of course there is nothing special about "3"- exactly the same thing can be done with an positive number "a", though the constant varies with a.) Using $3^x= e^{ln(3^x)}= e^{xln(3)}$ shows that the constant is ln(3).