7. Differential of a function - Mathematics for Business

7.1Implicit differentiation¶

In high-school, we learned about the explicit and implicit equation of line:

explicit form $\displaystyle \quad \dots \quad y = 2x+1$
implicit form $\displaystyle \quad \dots \quad 2x-y+1 = 0$

The difference between these two forms is that in explicit form we know the formula by which $y$ is given, while in implicit form we don’t immediately know that formula (although we can easily compute it). This same method can be used for more general functions.

Sometimes, the implict form of a function can be such that we can’t explicitly express $y$ from the given equation. Nonetheless, we know how to compute the derivative of implicitly given function, as the following problems show.

7.2Logarithmic differentiation¶

So far, we have learned how to find the derivatives of products, quotients, compositions and implicitly given functions. However, there are still some functions that we don’t (yet) know how to differentiate. In this section, we will learn a new method that we’ll be able to use to differentiate some functions that we didn’t know how to differentiate before.

7.3Differential of a function¶

Remember that the derivative of the function $f$ is defined as the limit

f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}.

If $h$ is very very close to 0, then it follows that

f(x+h)-f(x) \approx f'(x) \cdot h.

This approximation is very important to us, so we want to give special names to the quantities above:

small displacement $\dots dx = h$
exact change $\dots \Delta f(x_0) = f(x_0 + dx)-f(x_0)$
the differential $\dots df(x_0) = f'(x_0) \cdot dx.$

Therefore, the equation above tells us that

\Delta f(x_0) \approx df(x_0).

In other words, the exact change in the value of the function can be approximated by the differential.

Of course, the approximation can be accurate or not, so we need a way of measuring the error we make when using the differential to approximate the exact change:

absolute error $\displaystyle \dots \lvert \Delta f(x_0) - df(x_0)\rvert$
relative error $\displaystyle \dots \frac{\lvert \Delta f(x_0) - df(x_0)\rvert}{\Delta f(x_0)} \cdot 100\%$

Solution to Problem 7.9

If we denote by $A_0 = 14 600$ the initial amount of money that is being invested into advertising, then the first approach might be the to find the value of $A$ such that

R(A) = 1.03 \cdot R(A_0).

Since $R(A_0) \approx 1158.81,$ the equation that we need to solve is the following:

\ln(A) \cdot \sqrt{A + 5} = 1193.57

This equation is very complicated and we don’t know how to solve it using any methods that we know of. Therefore, we need to approach this problem in some other way.

Using the fact that the differential approximates the exact change, we have that

\Delta R(A_0) \approx R'(A_0) \cdot dA,

where $A_0 = 14600$ is the initial amount of money being invested in advertising. If the business wants to increase their revenue by $3\%,$ then $\Delta R(A_0) = 0.03R(A_0).$ So, from the approximation above, the additional amount of money that needs to be invested in advertising is

dA \approx \frac{\Delta R(A_0)}{R'(A_0)} = \frac{0.03R(A_0)}{R'(A_0)} = 725.02

Therefore, the bussiness needs to invest around 725 euros more in advertising in order to achieve their goal.

If we would like to check whether or not our approximation was good, we can compute the revenue if the bussines invests $A = 14600 + 725,02 = 15325.02$ euros in advertising:

R(15325,02) = 1193.23.

Since the $3\%$ increase of the initial revenue is equal to $1.03R(A_0) = 1193.57,$ we see that our approximation is reasonably good!

Solution to Problem 7.10

If we would like to know how the given change in the interst rate would affect the portfolio’s value, we would need to find $P(3.75).$ Plugging $r = 3,75$ into the given equation, we obtain

\frac{9}{16}P + P^4 = \frac{3e^{15}}{4} + 16.

Again, this is a rather complicated equation that we don’t know how to solve so we need to change our approach.

Because the differential approximates the exact change, we know that

\Delta P(r_0) \approx P'(r_0) \cdot dr,

where $r_0 = 3$ is the initial interest rate. In our case, we know that $dr = 0.75,$ and we want to compute $\Delta P(r_0) = \Delta P(3).$ Hence, the only thing we are missing is the value of the derivative $P'(3).$ Since the function $P$ is given implicitly, we use implicit differentiation to compute the derivative:

\begin{split} (r-3)^2 \cdot P + P^4 &= e^{4r} \cdot (r-3) + 16 \quad /' \\ 2(r-3)P + (r-3)^2 P' + 4P^3P' &= 4e^{4r}(r-3) + e^{4r} \end{split}

Pluggin-in $r = 3,$ we get

4P(3)^3P'(3) = e^{12}.

Plugging-in $r = 3$ into the defining equation for $P$ , we get

P(3)^4 = 16 \quad / \sqrt[4]{} \implies P(3) = 2.

Hence, we get

4P(3)^3P'(3) = e^{12} \implies 32 P'(3) = e^{12} \implies P'(3) = \frac{e^{12}}{32}.

Finally, we can approximate the change in portfolio’s value using the differential:

\Delta P(3) \approx P'(3) \cdot dr = \frac{e^{12}}{32} \cdot 0.75 = 3814.57

Solution to Problem 7.11

If the initial market index is equal to $I_0 = 2,$ then the initial price of the stock is $P(2) = 2744.$

Our task is to find by how much can the market index change so that Marko’s function gives accurate values up to $5\%$ of the inital value that is equal to $P(2) = 2744.$ If we denote that $dI$ that displacement of the market index, we would need to solve the following inequality:

\begin{split} \lvert P(2+dI) - P(2) \rvert &\leq \frac{5}{100} \cdot P(2) \\ \iff \lvert (14+5dI)^{\sqrt{9+dI}} - 2744 \rvert &\leq 137.2 \end{split}

Of course, this inequality is far to complicated for us to solve, so we need to change our approach.

Because the differential approximates the exact change, we know that

\Delta P(2) \approx P'(2) \cdot dI.

Therefore, we have the following approximation

\begin{split} \lvert P'(2) \cdot dI \rvert \approx \lvert \Delta P(2) \rvert &\leq \frac{1}{100} P(2). \\ \iff \lvert P'(2) \cdot dI \rvert &\leq 137.2 \end{split}

This inequality we know how to solve. First of all, we need to find the derivative $P'(2).$ Using logarithmic differentiation, we get

P'(I) = P(I) \cdot \left(\frac{\ln(5I+4)}{2\sqrt{I+7}} + \frac{5\sqrt{I+7}}{5I+4} \right)

If $I = 2,$ then we have

P'(2) = 2744 \cdot \left( \frac{\ln(14)}{6} + \frac{15}{14}\right).

Now, the problem boils down to a simple inequality

\begin{split} \lvert P'(2) \cdot dI \rvert &\leq 137.2 \\ \implies \lvert dI \rvert &\leq \frac{137.2}{P'(2)} = 0.033 \end{split}

Therefore, we have estimated that if the market index changes up to $\pm 0.033,$ accuracy of Marko’s model will be withing $5\%$ of the actual stock price.

In order to see if our estimates are accurate or not, let’s compute the exact change in portfolio value given that the market index changes by at most $\pm 0.03$ :

\begin{split} \lvert P(2,03)-P(2) \rvert & = 126.902 < 137.2 \\ \lvert P(1.97) - P(2) \rvert & = 121.972 < 137.2 \end{split}

Therefore, we see that our estimates are accurate, i.e. they ensure that the error made using Marko’s model remain under the given threshold.

Part 2. Functions of one variable

6. Derivative of a function

Part 2. Functions of one variable

8. Optimization of functions of one variable