(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: so ln(y)= x ln(3). Differentiate both sides with respect to x:
so .
Going back to the definition of derivative, the difference quotient would be . So the derivative is given by
Notice that the first part of that, does not have an "h" while the limit does not involve "x". That means that the derivative of is just 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 shows that the constant is ln(3).