(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!
(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!
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.)
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).