Get confused when differentiating exponentials? This article will explain how to differentiate and solve equations involving exponentials.
In this article, we will discuss how to differentiate exponentials, and apply chain rule to solve more complex exponentials.
Mastering this concept will give you a better understanding of the rate at which viruses spread and a better understanding of whether Australia is successfully combatting COVID-19!
NESA requires students to demonstrate proficiency in the following Syllabus outcomes:
E1.3 The exponential function and natural logarithms
E1.4 Graphs and applications of exponential and logarithmic functions
Students should be familiar with index laws.
Students should also be familiar with the concept of differentiation, and how to apply the product, quotient and chain rules.
Students can refresh their knowledge of differentiation in our Year 11 Guide Article: Calculus.
In calculus, when dealing with exponential functions, a common base to use is \(?\) (Euler’s number), where \(? \approx 2.71828\). This function \(e^x\) is called the exponential function.
The exponential function \(e^x\) is quite special as the derivative of the exponential function is equal to the function itself.
\(\frac{d}{dx} (e^x )= e^x\)
By applying chain rule, other standard forms for differentiation include:
\begin{align*} \frac{d}{dx}e^{ax+b} = ae^{ax+b} \\
\frac{d}{dx}(e^u ) = e^u \frac{du}{dx} \end{align*}
Find the gradient of the tangent of \(y=2e^{x^2+x}\)
This question follows the form, \(\frac{d}{dx}e^u=e^u \frac{du}{dx}\) where \(u=x^2+x\).
\(\frac{du}{dx}=2x+1\), using our basic differentiation knowledge.
Hence \(y’=2(2x+1)e^{x^2+x}\)
To derive \(a^x\), we must use \(\frac{d}{dx}e^x= e^x\).
\(a^x= e^{ln(a^x}\) (exponential and logarithms are inverse functions)
\(=e^{xln(a)}\) (property of logarithm)
Therefore,
\(\frac{d}{dx}a^x= ln(a)e^{xln(a)}=ln(a)a^x\)
An application of exponentials are finding the rate at which a quantity changes at a particular time. It is best to examine this through an example.
A population \(P\) is growing at a rate which is given by the formula \(P=1000e^{0.2t}\), where \(t\) is the number of days.
(a) Find the rate at which the population is changing and show that \(\frac{dP}{dt}=200e^{0.2t}\)
(b) Find the average rate of change over the first \(5\) days.
(c) When is the population increasing by \(1000\) per day?
(a) The rate the population is changing is the change of population over time, and hence equals to \(\frac{dP}{dt}\).
\(\frac{dP}{dt}=1000 \times 0.2e^{0.2t}=200e^{0.2t}\)
(b)
At \(t=0, P_0=1000\)
At \(t= 5, P_5=1000e\)
\begin{align*}
Hence \ average \ rate = \frac{P_5 – P_0}{t_5-t_0} \\
\cong \frac{1718.28}{5} \\
\cong 344 people \ a \ day \\
\end{align*}
(c)
\begin{align*}
\frac{dP}{dt}=1000 \\
200e^{0.2t}=1000 \\
e^{\frac{t}{5}}=5 \\
\frac{t}{5}=ln(5) \\
t=5ln(5) \\
\cong 8 \ days
\end{align*}
A normal exponential \(y=e^x\) graph is shown below, and hence \(lim_{x→-\infty}e^x=0\) and there is a horizontal asymptote at \(y=0\).
We can use the usual transformations to change the exponential graphs.
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1. Find the derivative of \(3e^{2x+1}\)
2. Find the equation of the tangent at \(f(x)=xe^{2x+7}\) at \(x=1\)
3. A radioactive sample with mass \(M\) decays at such a rate that \(M=M_0 e^{-0.2t}\), where \(M_0\) is the original mass and \(t\) is the time in years.
(a) How long until half the sample is gone?
(b) Find the rate at which the sample is decaying and prove that \(\frac{dM}{dt}=-0.2 \times M\)
(c) When will the rate of decay be \(2%\) of the original mass \(M_0\) per year?
4. Graph \(y=3e^{-(x+1)}+1\)
1. Since \(\frac{d}{dx}e^{ax+b}= ae^{ax+b}\), the derivative of \(e^{2x+1}\) is \(2e^{2x+1}\), then the derivative of \(3e^{2x+1}\), is \(3\times 2e^{2x+1}= 6e^{2x+1}\)
2. We can use the product rule to solve this where \(u=x\) and \(v= e^{2x+7}\).
\(u’=1\) and \(v’= 2e^{2x+7}\)
Using the product rule, \(f'(x)=2xe^{2x+7}+e^{2x+7}\)
Hence, the gradient at \(x=1\) is \(f’ (1)=2e^9+e^9=3e^9\)
To find an exact point that this tangent goes through, \(f(1)=e^9\). Hence, the tangent has a gradient of \(3e^9\) and goes through the point \((1,e^9)\).
By substituting this into the point-gradient form of a line, and rearranging, the solution is \(y=3e^9 \times – 2e^9\).
3.
(a) Since \(M_0\) is the original mass, then \(\frac{1}{2} M_0\) is half the sample.
\begin{align*} \frac{1}{2} M_0=M_0 e^{-0.2t} \\
\frac{1}{2}=e^{-0.2t} \\
-0.2t=ln(\frac{1}{2} \\
t=5ln(2) \approx 3.5 \ years \end{align*}
(b) \(\frac{dM}{dt}=-0.2 \times M_0 e^{-0.2t}=-0.2M\)
(c) When \(\frac{dM}{dt}=-\frac{1}{50}M_0\),
\begin{align*} -\frac{1}{50}M_0=-0.2 \times M_0 e^{-0.2t} \\
\frac{1}{10}=e^{-0.2t} \\
t=5 ln(10) \approx 11.5 \ years \end{align*}
4. We can graph this using multiple transformations:
Shift the graph \(y=e^x\) graph to the left by \(1\) unit to get \(y=e^{x+1}\)
Then reflect on the \(y\) – axis, to get \(y=e^{-(x+1)}\)
Apply a vertical stretch by \(3\), to get \(y=3e^{-(x+1)}\)
Shift the graph up \(1\) unit, to get \(y=3e^{-(x+1)}+1\)
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