(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

...

Please help me! Thank you!

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- Jun 27th 2014, 01:32 AMpumbaa213Need 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

...

Please help me! Thank you! - Jun 27th 2014, 03:31 AMpumbaa213Re: Need help on this question. thanks guys
its okay guys. I managed to solve it (finally!)

Heres my answer:

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, 05:55 AMHallsofIvyRe: Need help on this question. thanks guys
Another way to do this: $\displaystyle y= 3^x$ so ln(y)= x ln(3). Differentiate both sides with respect to x:

$\displaystyle \frac{1}{y}\frac{dy}{dx}= ln(3)$ so $\displaystyle \frac{dy}{dx}= y ln(3)= 3^x ln(3)$.

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

Notice that the first part of that, $\displaystyle 3^x$ does not have an "h" while the limit does not involve "x". That means that the derivative of $\displaystyle 3^x$ is just $\displaystyle 3^x$ 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 $\displaystyle 3^x= e^{ln(3^x)}= e^{xln(3)}$ shows that the constant is ln(3).