11. Differentiation

iGCSE (2021 Edition)

Lesson

We have just explored the differentiation of exponential functions. As with the derivatives of other functions which we have seen, we can use this to find the equation of tangents and normals to exponential functions.

Standard forms for differentiating exponential functions

$\frac{d}{dx}e^x$ddxex |
$=$= | $e^x$ex |

$\frac{d}{dx}e^{ax}$ddxeax |
$=$= | $ae^{ax}$aeax |

$\frac{d}{dx}e^{f\left(x\right)}$ddxef(x) |
$=$= | $f'\left(x\right)e^{f\left(x\right)}$f′(x)ef(x) |

When finding equations of tangents and normals, remember that the gradient of the tangent at any point on a function is given by the first derivative of the function. The following formulas are also useful:

Formulas for tangents and normals

Given the gradient $m$`m` and a point $\left(x_1,y_1\right)$(`x`1,`y`1), the equation of a line can be found using the point gradient formula:

$y-y_1=m\left(x-x_1\right)$`y`−`y`1=`m`(`x`−`x`1)

Given the gradient of the tangent $m_1$`m`1 at a point, the gradient of the normal $m_2$`m`2 is:

$m_2=\frac{-1}{m_1}$`m`2=−1`m`1

Find the equation of the tangent and the normal to the curve given by $y=e^{x^2-4x}$`y`=`e``x`2−4`x` at the point where $x=4$`x`=4.

**Think:** To find the equation of the tangent, we will use the point gradient formula. For this we need to find the $y$`y` coordinate on the function for $x=4$`x`=4.

**Do:**

$y$y |
$=$= | $e^{x^2-4x}$ex2−4x |

$y$y |
$=$= | $e^{4^2-16}$e42−16 |

$y$y |
$=$= | $e^0$e0 |

$y$y |
$=$= | $1$1 |

Next, we will determine the gradient function and evaluate the gradient at $x=4$`x`=4. We do this by finding the first derivative of the function as follows:

$y$y |
$=$= | $e^{x^2-4x}$ex2−4x |

$\frac{dy}{dx}$dydx |
$=$= | $(2x-4)e^{x^2-4x}$(2x−4)ex2−4x |

For $x=4$`x`=4:

$\frac{dy}{dx}$dydx |
$=$= | $(2\times4-4)e^{4^2-16}$(2×4−4)e42−16 |

$\frac{dy}{dx}$dydx |
$=$= | $4$4 |

Using the point gradient formula, with the point $\left(4,1\right)$(4,1) and gradient of $4$4 we can find the equation of the tangent as follows:

$y-y_1$y−y1 |
$=$= | $m\left(x-x_1\right)$m(x−x1) |

$y-1$y−1 |
$=$= | $4(x-4)$4(x−4) |

$y-1$y−1 |
$=$= | $4x-16$4x−16 |

$y$y |
$=$= | $4x-15$4x−15 |

To find the equation of the normal at the same point we first find the gradient. We know the gradient of the tangent $m_1=4$`m`1=4, hence the gradient of the normal $m_2$`m`2 is given by:

$m_2$m2 |
$=$= | $-\frac{1}{m_1}$−1m1 |

$m_2$m2 |
$=$= | $-\frac{1}{4}$−14 |

Using the point gradient formula, with the point $\left(4,1\right)$(4,1) and gradient of $-\frac{1}{4}$−14 we can find the equation of the tangent as follows:

$y-y_1$y−y1 |
$=$= | $m\left(x-x_1\right)$m(x−x1) |

$y-1$y−1 |
$=$= | $-\frac{1}{4}(x-4)$−14(x−4) |

$y-1$y−1 |
$=$= | $-\frac{x}{4}+1$−x4+1 |

$y$y |
$=$= | $-\frac{x}{4}+2$−x4+2 |

Find the equation of the tangent to the curve $f\left(x\right)=2e^x$`f`(`x`)=2`e``x` at the point where it crosses the $y$`y`-axis.

Express the equation in the form $y=mx+c$`y`=`m``x`+`c`.

The first and second derivatives of exponential functions can be used to determine concavity, and regions where an exponential function is increasing or decreasing. Additionally, any stationary points and points of inflection can be found. Using these tools we can sketch functions involving exponentials.

Sketch the function $f\left(x\right)=\frac{1}{2}e^{-x^2}$`f`(`x`)=12`e`−`x`2 showing any intercepts and stationary points.

**Think:** This function involves an even index and therefore it would be worth checking if it is an even function. If it is an even function we will be able to exploit the symmetrical nature.

**Do:** To check if it is an even function we test if $f\left(x\right)=f\left(-x\right)$`f`(`x`)=`f`(−`x`):

$f\left(x\right)$f(x) |
$=$= | $\frac{1}{2}e^{-x^2}$12e−x2 |

$f\left(-x\right)$f(−x) |
$=$= | $\frac{1}{2}e^{-(-x)^2}$12e−(−x)2 |

$f\left(-x\right)$f(−x); |
$=$= | $\frac{1}{2}e^{-x^2}$12e−x2 |

$f\left(-x\right)$f(−x) |
$=$= | $f\left(x\right)$f(x) |

Therefore, it is an even function and symmetrical about the $y$`y`-axis.

To find the $y$`y`-intercept, we make $x=0$`x`=0:

$f\left(0\right)$f(0) |
$=$= | $\frac{1}{2}e^{-0^2}$12e−02 |

$f\left(0\right)$f(0) |
$=$= | $\frac{1}{2}$12 |

To find any $x$`x`-intercepts, we make $y=0$`y`=0:

$0$0 | $=$= | $\frac{1}{2}e^{-x^2}$12e−x2 |

There are no solutions to this equation and hence, there are no $x$`x`-intercepts.

To check for any stationary points we need to find the first derivative and make it equal to $0$0:

$f\left(x\right)$f(x) |
$=$= | $\frac{1}{2}e^{-x^2}$12e−x2 |

$f'\left(x\right)$f′(x) |
$=$= | $-2x\frac{1}{2}e^{-x^2}$−2x12e−x2 |

$f'\left(x\right)$f′(x) |
$=$= | $-xe^{-x^2}$−xe−x2 |

$0$0 | $=$= | $-xe^{-x^2}$−xe−x2 |

$x$x |
$=$= | $0$0 |

Therefore, there is a stationary point at $x=0$`x`=0.

We will find the second derivative and use it to determine the nature of the stationary point and check for any points of inflection:

$f''\left(x\right)$f′′(x) |
$=$= | $-e^{-x^2}+-2xe^{-x^2}(-x)$−e−x2+−2xe−x2(−x) |

$f''\left(x\right)$f′′(x) |
$=$= | $-e^{-x^2}+2x^2e^{-x^2}$−e−x2+2x2e−x2 |

$f''\left(x\right)$f′′(x) |
$=$= | $e^{-x^2}\left(2x^2-1\right)$e−x2(2x2−1) |

At $x=0$`x`=0:

$f''\left(0\right)$f′′(0) |
$=$= | $e^{-2(0)^2}\left(2(0)^2-1\right)$e−2(0)2(2(0)2−1) |

$f''\left(0\right)$f′′(0) |
$=$= | $-1$−1 |

Therefore, there is a local maximum at $x=0$`x`=0 because $f''\left(x\right)<0$`f`′′(`x`)<0 and hence the curve is concave down at this point.

Setting $f''\left(x\right)=0$`f`′′(`x`)=0, we we can check for any inflection points:

$0$0 | $=$= | $e^{-x^2}\left(2x^2-1\right)$e−x2(2x2−1) |

$0$0 | $=$= | $2x^2-1$2x2−1 |

$2x^2$2x2 |
$=$= | $1$1 |

$x^2$x2 |
$=$= | $\frac{1}{2}$12 |

$x$x |
$=$= | $\pm\frac{1}{\sqrt{2}}$±1√2 |

Therefore, there are possible points of inflection at $x=\pm\frac{1}{\sqrt{2}}$`x`=±1√2.

We could check either the first derivative for the sign of the gradient or the second derivative for a change in concavity, to the left and right of these values, to determine if they are points of inflection. We will look at both below, but in practice, you would choose the simplest option.

x | $-1$−1 | $-\frac{1}{\sqrt{2}}$−1√2 | $0$0 | $\frac{1}{\sqrt{2}}$1√2 | $1$1 |
---|---|---|---|---|---|

$f'\left(x\right)$f′(x) |
$0.37$0.37 | $0.43$0.43 | $0$0 | $-0.43$−0.43 | $-0.37$−0.37 |

$f''\left(x\right)$f′′(x) |
$0.37$0.37 | $0$0 | $-1$−1 | $0$0 | $0.37$0.37 |

As the gradient remains the same and there is a change in concavity at both $x=\pm\frac{1}{\sqrt{2}}$`x`=±1√2 we can classify the points as points of inflection.

Finally, we will consider the behaviour of the function as $x\rightarrow\pm\infty$`x`→±∞, first, we will rearrange $f\left(x\right)$`f`(`x`) to make the function easier to analyse:

$f\left(x\right)$f(x) |
$=$= | $\frac{1}{2}e^{-x^2}$12e−x2 |

$f\left(x\right)$f(x) |
$=$= | $\frac{1}{2e^{x^2}}$12ex2 |

As $x\rightarrow\infty$`x`→∞, the denominator of the function will become very large and positive, hence the function will approach zero from above:

As $x\rightarrow\infty$x→∞ |
$f\left(x\right)\rightarrow0$f(x)→0 |

We could use symmetry and say the behaviour will be the same as $x\rightarrow-\infty$`x`→−∞. Let's consider $x\rightarrow-\infty$`x`→−∞ anyway. The denominator of the function will become very large and positive, hence the function will approach zero from above:

As $x\rightarrow-\infty$x→−∞ |
$f\left(x\right)\rightarrow0$f(x)→0 |

All that is left is to sketch the function showing all the important points:

Consider the function $f\left(x\right)=3-e^{-x}$`f`(`x`)=3−`e`−`x`.

Determine $f\left(0\right)$

`f`(0).Determine $f'\left(0\right)$

`f`′(0).Which of the following statements is true?

$f'\left(x\right)<0$

`f`′(`x`)<0 for $x\ge0$`x`≥0A$f'\left(x\right)<0$

`f`′(`x`)<0 for all real $x$`x`.B$f'\left(x\right)>0$

`f`′(`x`)>0 for all real $x$`x`.C$f\left(x\right)>0$

`f`(`x`)>0 for all real $x$`x`.D$f'\left(x\right)<0$

`f`′(`x`)<0 for $x\ge0$`x`≥0A$f'\left(x\right)<0$

`f`′(`x`)<0 for all real $x$`x`.B$f'\left(x\right)>0$

`f`′(`x`)>0 for all real $x$`x`.C$f\left(x\right)>0$

`f`(`x`)>0 for all real $x$`x`.DDetermine the value of $\lim_{x\to\infty}f\left(x\right)$lim

`x`→∞`f`(`x`).

Consider the function $f\left(x\right)=4e^{-x^2}$`f`(`x`)=4`e`−`x`2.

Determine $f'\left(x\right)$

`f`′(`x`).Determine the values of $x$

`x`for which $f'\left(x\right)=0$`f`′(`x`)=0.Determine the values of $x$

`x`for which $f'\left(x\right)>0$`f`′(`x`)>0.Determine the values of $x$

`x`for which $f'\left(x\right)<0$`f`′(`x`)<0.Determine the value of $\lim_{x\to\infty}f\left(x\right)$lim

`x`→∞`f`(`x`).Determine the value of $\lim_{x\to-\infty}f\left(x\right)$lim

`x`→−∞`f`(`x`).Which of the following is the graph of $f\left(x\right)$

`f`(`x`)?Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...ALoading Graph...BLoading Graph...C

An exponential function is the appropriate model to use when a quantity is increasing or decreasing at a rate that depends on the quantity present.

For example, in the final rounds of a sports competition, the number of competing teams is halved at every stage. Thus, if $16$16 teams reached the first semifinal, there would be $8$8 in the second semifinal, and so on. The number of teams playing drops from $16$16 to $8$8 and then to $4$4 and finally, $2$2 and we see that the reduction in the number of teams playing at each stage depends on the number in the previous round.

This is an example of exponential decay. The rate of decrease gets progressively smaller. Many processes show the opposite pattern and exhibit exponential growth. In this, the rate of increase increases progressively.

Remember!

If we are looking at an exponential function of the form $y=ab^x$`y`=`a``b``x`, then

- $b$
`b`is tells us whether the function is growing (increasing) or decaying (decreasing)- If $b>1$
`b`>1, it is growth - If $00<
`b`<1, it is decay

- If $b>1$
- $a$
`a`is the initial value, when $x=0$`x`=0

One of the most common applications is where there is a percentage growth or a percentage decay. For example the amount in a savings account or the remaining balance on a loan respectively. Let's look at the general formula for these scenarios.

A particular sports convertible costs $\$48000$$48000. It loses $18%$18% of its value per year. We want to find its value for the next $4$4 years. Let's use a table to do so.

Year | Amount Lost | Value ($\$$$) | Expression for Value |
---|---|---|---|

$0$0 | n/a | $48000$48000 | $48000$48000 |

$1$1 | $0.18\times48000=8640$0.18×48000=8640 | $48000-8640=39360$48000−8640=39360 | $48000-48000\times0.18=48000\left(1-0.18\right)$48000−48000×0.18=48000(1−0.18) |

$2$2 | $0.18\times39360=7084.80$0.18×39360=7084.80 | $39360-7084.80=32275.20$39360−7084.80=32275.20 | $48000\left(1-0.18\right)^2$48000(1−0.18)2 |

$3$3 | $0.18\times32275.20=5809.54$0.18×32275.20=5809.54 | $32275.20-5809.54=26465.66$32275.20−5809.54=26465.66 | $48000\left(1-0.18\right)^3$48000(1−0.18)3 |

$4$4 | $0.18\times26465.66=4763.82$0.18×26465.66=4763.82 | $26465.66-4763.82=21701.84$26465.66−4763.82=21701.84 | $48000\left(1-0.18\right)^4$48000(1−0.18)4 |

Rather than continuing to work through each year in this way, we could approach the problem differently by looking at the column Expression for value.

Notice that multiplying by $18%$18% and then subtracting that from the current value is the same as multiplying by $\left(1-0.18\right)=0.82$(1−0.18)=0.82. Since we are doing this repeatedly, we can make a formula.

In fact, for exponential growth and decay by a percentage increase or decrease, we can use the formula below:

Exponential growth and decay formula

$A=P\left(1+r\right)^n$`A`=`P`(1+`r`)`n`

where:

- $A$
`A`is the resulting amount after the growth/decay - $P$
`P`is the initial amount - $r$
`r`is the rate of change as a decimal and indicates growth if $r>0$`r`>0, and decay if $r<0$`r`<0 - $n$
`n`is the number of time periods that have passed from the start

The frequency of $r$`r` and $n$`n` must match! So if $r$`r` is yearly, then $n$`n` must be in years, but if $r$`r` is a weekly rate, then $n$`n` must be in weeks.

Justine bought an oil painting for $\$1600$$1600 whose value is given by the formula $A=1600\times0.94^t$`A`=1600×0.94`t`, and $t$`t` is the number of years passed.

**(a)** What's the annual depreciation rate?

**Think: **What's $r$`r` in this case?

**Do:**

$1600\times0.94^t=1600\left(1-0.06\right)^t$1600×0.94`t`=1600(1−0.06)`t`

so the annual depreciation rate is $0.06=6%$0.06=6%.

**(b)** How much would the painting be worth in 10 years, round to the nearest dollar?

**Think:** What parameters do we have, which one are we looking for. It can be helpful to list values before jumping into the formula. For the equation, $A=1600\times0.94^t$`A`=1600×0.94`t`, we are given $t=10$`t`=10 and are looking for $A$`A`.

**Do:**

$A$A |
$=$= | $1600\times0.94^t$1600×0.94t |

$=$= | $1600\times0.94^{10}$1600×0.9410 | |

$=$= | $861.7841826$861.7841826 |

So the painting would be worth $\$862$$862 in $10$10 years.

A certain radioactive isotope decays in such a way that after $175$175 years only half of the original quantity of the isotope remains. Suppose $10$10 kg of the substance existed initially. Create an equation for the amount of the isotope left after $t$`t` years.

**Think:** When given a worded problem, it helps to list what we were given and what we are looking for.

$A$`A`: This will be an unknown based on the number of years, $t$`t`

$P$`P`: We initially have $10$10 kg, so $P=10$`P`=10

$r$`r`: This is the rate of decay, which we can say is $50%$50% if we are looking at the half-life

$n$`n`: This will be filled in based on the time, we need to remember that the rate is in groups of $175$175 years, so $n$`n` must be in groups of 175 years

**Do**:

After $175$175 years, only $5$5 kg will be left and then after a further $175$175 years, only $2.5$2.5 kg will be left, and so on. After $n$`n` groups of $175$175 years, we have that:

$A=10\times\left(\frac{1}{2}\right)^n$`A`=10×(12)`n`

The number of years that have elapsed since the beginning of this experiment must be $t=175n$`t`=175`n`. So, we can put $n=\frac{t}{175}$`n`=`t`175 into the formula so that we no longer have to convert the time elapsed into groups of $175$175 years but can use just *years *instead.

Thus, we arrive at a formula for the amount remaining after $t$`t` years:

$A=10\times\left(\frac{1}{2}\right)^{\frac{t}{175}}$`A`=10×(12)`t`175

**Reflect**: How could we use this equation? How much would be left after $1000$1000 years?

The population of a particular mining town increased $160%$160% in $9$9 years, from $5100$5100 in 2004 to $13260$13260 in 2013. Assuming that the population increased at a constant annual rate, answer the following.

Find an expression for $A$

`A`, the size of the population $y$`y`years after 2004. Write the expression such that it includes the annual rate of growth, correct to four decimal places.Hence state the annual rate of growth. Give the rate as a percentage correct to two decimal places.

Consider the function $f\left(t\right)=\frac{8}{7}\left(\frac{3}{8}\right)^t$`f`(`t`)=87(38)`t`, where $t$`t` represents time.

What is the initial value of the function?

Express the function in the form $f\left(t\right)=\frac{8}{7}\left(1-r\right)^t$

`f`(`t`)=87(1−`r`)`t`, where $r$`r`is a decimal.Does the function represent growth or decay of an amount over time?

decay

Agrowth

Bdecay

Agrowth

BWhat is the rate of decay per time period? Give the rate as a percentage.

Use the derivatives of the standard functions e^x, ln x, together with constant multiples, sums and composite functions.