Tag Archives: Calculus

January 9, 2025 · 9:40 am

Deriving the Chain Rule for Differentiation

How to differentiate something in the form y=[f(x)]^n

For example, y=(3x^2-2x+6)^5, we could expand the expression, but the Chain Rule provides a quick and easy method.

Differentiate y=[f(x)]^n

Let u=f(x), then y=u^n

We want to find \frac{dy}{dx}, but \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}

They’re not fractions, but limits of fractions, but they work like fractions.

\frac{du}{dx}=f'(x) and \frac{dy}{du}=nu^{n-1}

Therefore, \frac{dy}{dx}=f'(x)\times nu^{n-1}

Replace u with f(x)

(1)   \begin{equation*}\frac{dy}{dx}=n[f(x)]^{n-1}f'(x)\end{equation*}

What about a function in the form y=f(g(x))?

We’re going to follow the same process.

Let u=g(x), then y=f(u)

\frac{du}{dx}=g'(x) and \frac{dy}{du}=f'(u)

Therefore \frac{dy}{dx}=f'(u)g'(x)

(2)   \begin{equation*}\frac{dy}{dx}=f'(g(x))g'(x) \end{equation*}

Equations 1 and 2 are versions of the Chain Rule.

Example

Find the derivative of y=(3x^2-2x+6)^5

    \begin{equation*}\frac{dy}{dx}=5(3x^2-2x+6)^4\times (6x-2)\end{equation}

    \begin{equation*}\frac{dy}{dx}=5(6x-2)(3x^2-2x+6)^4\end{equation}

    \begin{equation*}\frac{dy}{dx}=10(3x-1)(3x^2-2x+6)\end{equation}

Next time we are going to look at the Product Rule.

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November 17, 2024 · 5:23 am

Integration by Parts using the Tabular Method

(1)   \begin{equation*}\int_0^1x^2e^xdx\end{equation*}

I am going to do this integral in two ways; the traditional method and the tabular method.

Traditional Method

Remember \int{u dv}=u\times v-\int{v du}

Let u=x^2 and dv=e^x

Then du=2x and v=\int{e^x dx}=e^x

    \begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-\int_0^1{e^x 2x dx}\end{equation}

Now we need to do integration by parts on \int_0^1{e^x 2x dx}

Let u=2x and dv=e^x

Then du=2 and v=e^x

    \begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-(2x\times e^x]_0^1-\int_0^1{e^x 2 dx})\end{equation}

    \begin{equation*}\int_0^1{x^2e^x dx}=x^2 \times e^x ]_0^1-(2x\times e^x]_0^1-(2e^x )]_0^1)\end{equation}

    \begin{equation*}\int_0^1{x^2e^x dx}=e-(2e-(2e-2))\end{equation}

    \begin{equation*}\int_0^1{x^2e^x dx}=e+2\end{equation}

Tabular Integration

Similar to before, select a u and a dv, u=x^2 and dv=e^x

SignD(ifferentiate)I(ntegrate)
+x^2e^x
–2xe^x
+2e^x
–0e^x

Stop when the differentiating column reaches zero.

Then we multiply diagonally

(+x^2)(e^x)+(-2x)(e^x)+(+2e^x)

=x^2e^x-2xe^x+2e^x]_0^1

=e-2e+2e-(0-0-2)

=e+2

It is only worth using this method if integration by parts is required more than once. Also, the u has to eventually differentiate to 0.

Let’s try another one

(2)   \begin{equation*}\int{x^3cos(2x) dx}\end{equation*}

Let u=x^3 and dv=cos(2x)

SignDI
+x^3cos(2x)
–3x^2\frac{1}{2}sin(2x)
+6x-\frac{1}{4}cos(2x)
–6-\frac{1}{8}sin(2x)
+0\frac{1}{16}cos(2x)

\int{x^3cos(2x) dx}=(x^3)(\frac{1}{2}sin(2x))+(-3x^2)(-\frac{1}{4}cos(2x))+(6x)(-\frac{1}{8}sin(2x))+(-6)(\frac{1}{16}cos(2x))+c

=\frac{x^3}{2}sin(2x)+\frac{3x^2}{4}cos(2x)-\frac{3x}{4}sin(2x)-\frac{3}{8}cos(2x)+c

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October 28, 2024 · 6:51 am

Related Rates Question

A boat is moving towards the beach line at 3 metres/minute. On the boat is a rotating light, revolving at 4 revolutions/minute clockwise, as observed from the beach. There is a long straight wall on the beach line, as the boat approaches the beach, the light moves along the wall. Let x equal the displacement of the light from the point O on the wall, which faces the boat directly. See the diagram below.
Determine the velocity, in metres/minute, of the light when x=5 metres, and the distance of the boat from the beach D is 12 metres.

Mathematics Specialist Semester 2 Exam 2018


The light is rotating at 4 revolutions/minute, which means

    \begin{equation*}\frac{d\theta}{dt}=4\times\pi\end{equation}

We want to find \frac{dx}{dt} and we know \frac{d\theta}{dt} and \frac{dD}{dt}.

We need to find an equation connecting x, \theta, and D.

    \begin{equation*}tan (\theta)=\frac{x}{D}\end{equation}

Differentiate (implicitly) with respect to time.

    \begin{equation*}sec^2(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}D-\frac{dD}{dt}x}{D^2}\end{equation}

Now we know x=5, and D=12, using pythagoras we can calulate the hypotenuse.


h=\sqrt{12^2+3^2}=13
sec(\theta)=\frac{13}{12}

    \begin{equation*}sec^2(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}D-\frac{dD}{dt}x}{D^2}\end{equation}

    \begin{equation*}(\frac{13}{12})^2\times 4\pi=\frac{\frac{dx}{dt}12-3\times 5}{12^2}\end{equation}

    \begin{equation*}169\times4\pi=12\frac{dx}{dt}-15\end{equation}

    \begin{equation*}676\pi+15=12\frac{dx}{dt}\end{equation}

    \begin{equation*}\frac{dx}{dt}=178.2\end{equation}

The velocity of the light is 178.2 m/minute.

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September 24, 2024 · 3:30 am

Optimisation

An optimisation question from the 2019 ATAR Mathematics Methods exam.

I always like optimisation questions. There is a nice process to follow:

  • Find the function to optimise (in terms of one variable).
  • Find the stationary points.
  • Find the nature of the stationary points.
  • Find the maximum or minimum.
(a) Volume of the cylinder V=\pi r^2h
42=2r+h
h=42-2r
\therefore V_C=\pi r^2(42-2r)
Volume of spherical decorations V_S=\frac{4}{3}\pi( r_s)^3 where r_s=\frac{r}{3}
V_S=\frac{4\pi r^3}{81}
Volume unused space V=\pi r^2(42-2r)-20(\frac{4\pi r^3}{81})
V=2\pi (21r^2-r^3-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{81r^3}{81}-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{121r^3}{81})

(b) V=2\pi (21r^2-\frac{121r^3}{81})
\frac{dV}{dr}=2\pi (42r-\frac{121r^2}{27})
\frac{dV}{dr}=0
0=42r-\frac{121r^2}{27}
0=r(42-\frac{121r}{27})
r=0 or r=\frac{1134}{121}=9.372

\frac{d^2V}{dr^2}=2\pi (42-\frac{242r}{27})
(\frac{d^2V}{dr^2})_{|r=9.372}=-42
\therefore r=9.372 is a maximum.

Dimensions of the vase, internal diameter=18.7cm internal height=23.3cm

(c) Maximum volume of empty space =2\pi (21r^2-\frac{121r^3}{81})=3863.08cm^3
Volume of one sphere =\frac{4}{3}\pi r^3=3448.03cm^3

There is enough unused space for one extra decoration, but it would depend on how they are packed.

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