Question

What is the rate of increase in this function: y=20(1.31)^x?

Answer 1

Remember the following properties of the derivatives:

Derivative of an exponential function:

`(d)/(dx)ke^x=ke^x`

Chain rule:

`(d)/(dx)f(u)=(d)/(du)f(u)\cdot(du)/(dx)`

Also, remember the following property of exponentials:

`a^x=e^(x\cdot\ln (a))`

Use these properties to find the derivative of the following function:

`f(x)=k\cdot a^x`

The function can be rewritten as:

`f(x)=k\cdot e^(x\cdot\ln (a))`

Let:

`u=x\cdot\ln (a)`

Then:

`\begin{gathered} (d)/(dx)k\cdot e^(x\cdot\ln (a))=(d)/(dx)k\cdot e^u \\ =(d)/(du)(k\cdot e^u)\cdot(du)/(dx) \\ =k\cdot e^u\cdot(d)/(dx)(x\cdot\ln (a)) \\ =k\cdot e^(x\cdot\ln (a))\cdot\ln (a) \\ =k\cdot a^x\cdot\ln (a) \end{gathered}`

In this case, we can see that k=20 and a=1.31:

`y=20\cdot1.31^x`

Then, the derivative of this function, is:

`y^(\prime)=20\cdot1.31^x\cdot\ln (1.31)`

Therefore, the rate of increase of the given function, is:

`y^(\prime)=20\cdot\ln (1.31)\cdot(1.31)^x`