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

User Mark Rowe
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1 Answer

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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

User Wingedsubmariner
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