Calc question?

Question

Estimate P'(0) if P(t)=200(1.05)^t. Please explain how to do this

Answer 1

P'(t) = 200 (1.05)^t . ln (1.05)
So P'(0) = 200 . 1 . ln (1.05)
Now, if the intent is to use Taylor series to expand this, you should know that ln (1+x) ~ x - x^2/2 + x^3/3 - ... for x small. So we can approximate this as
200 (1/20 - 1/800 + 1/24000)
= 10 - 0.25 + 0.008333... = 9.76 to 2 d.p.

Answer 2

Write P(t) as 200* e^(t*ln 1.05)
Then P'(t) = ln 1.05 * 200 * e^t
then P'(0) = ln 1.05 * 200 * e^0
Since e^0 = 1, this would be 200 ln 1.05. Punch this into your calculator for a decimal approximation.

Answer 3

I would probably convert the exponent expression into an exponential function because I know how to differentiate that!

Let k=ln[200(1.05)]

then P(t) = exp(kt)

P'(t) = kexp(kt)

Now just set t to zero
P'(t=0) = k

Answer 4

Find (P(0 + delta t) - P(0 - delta t))/(2*delta t)

P(0.01) = 200(1.05)^0.01 = 200.1
p(-0.01)= 200(1.05)^-0.01 = 199.9
p'(0) = (200.1-199.9)/0.02 = 10