Derivative question

Worth "y" kr for a "x" year old car decreases according to y=250000*0.9^x

determine and interpret

a) y(3)
y'(3)

So we have been doing derivatives for a while now and I'm starting to get hang of the basics of all the different ways to calculate and solve problems.

y(3) in this scenario is easy.

250000*0.9^3=182250

but how am I supposed to put in the prime y'(3)?

The answer on is supposed to be y'(3)=-19000kr
After 3 years the cars value decreases with the speed of 19000kr/year

kr is the currency we have in Sweden for money.

I would start by differentiating the original function

Y’(3) is just telling you to input 3 into the derivative. Once you have the derivative it will be the same as what you did for a.

This post was edited by brigadier on Mar 1 2019 09:12pm

You will need to use Implicit Differentiation to find the derivative.

First, take the natural log on both sides to utilize the power rule for logarithms.
Second, take the derivative with respect to x on both sides. This will have y still in the derivative, which you back substitute later on.
Third, solve for y' (or dy/dx).
Finally, replace y in the derivative with the original expression 250000*0.9^x.