### Differentiation Question #1

Given that $y = \frac{8}{x^3} - \frac{6}{x^2} + \frac{5}{2x}$, find the approximate percentage change in $y$ when $x$ increases from 2 by 2%.

### Random Sec 4 Differentiations

B6

$y = 3e^x + \frac{4}{e^x}$

$\frac{dy}{dx} = 3e^x - \frac{4}{e^x}$

$\frac{d^2y}{dx^2} = 3e^x + \frac{4}{e^x}$

let $\frac{dy}{dx} = 0$

$3e^x - \frac{4}{e^x} = 0$

$3e^{2x} = 4$

$2x = \mathrm{ln} \frac{4}{3}$

$x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$

Sub $x = \frac{1}{2} \mathrm{ln} \frac{4}{3}$ to $\frac{d^2y}{dx^2}$

$\frac{d^2y}{dx^2} > 0$ Thus, it is a min point.

C7

$y = \mathrm{ln} \frac{5-4x}{3+2x}$

$y = \mathrm{ln} (5-4x) - \mathrm{ln} (3+2x)$

$\frac{dy}{dx} = \frac{-4}{5-4x} - \frac{2}{3+2x}$

let $\frac{dy}{dx} = 0$

$\frac{-4}{5-4x} - \frac{2}{3+2x} = 0$

$\frac{-4}{5-4x} = \frac{2}{3+2x}$

$-4(3+2x) = 2(5-4x)$

$-12 - 8x = 10 - 8x$

$-12 = 10$ (NA).

There are no stationary points for this curve.

C8

$x = \frac{1}{3}e^{y(2x+5)}$

$\mathrm{ln}(3x) = y(2x+5)$

$\frac{\mathrm{ln}(3x)}{2x+5} = y$

$y = \frac{\mathrm{ln}(3x)}{2x+5}$

$\frac{dy}{dx} = \frac{\frac{1}{x}(2x+5) - \mathrm{ln}(3x) \times 2}{(2x+5)^2}$

Let $x = e^2$

$\frac{dy}{dx} = \frac{\frac{1}{e^2}(2e^2+5) - \mathrm{ln}(3e^2) \times 2}{(2e^2+5)^2}$

Evaluate with a calculator…

### A little reminder to students doing Calculus now

When $\frac{dy}{dx} = 0$, it implies we have a stationary point.

To determine the nature of the stationary point, we can do either the first derivative test or the second derivative.

The first derivative test:

Students should write the actual values of $\alpha^-, \alpha, \alpha^+$ and $\frac{dy}{dx}$ in the table.

We use this under these two situations:
1. $\frac{d^2y}{dx^2}$ is difficult to solve for, that is, $\frac{dy}{dx}$ is tough to be differentiated
2. $\frac{d^2y}{dx^2} = 0$

The second derivative test:

Other things students should take note is concavity and drawing of the derivative graph.

### Random Questions from 2016 Prelims #4

VJC P1 Q6

A curve has equation $y^2=4x$ and a line $l$ has equation $2x-y+1=0$ as shown above.

$B(b, 2\sqrt{b})$ is a fixed point on C and A is an arbitrary point on $l$. State the geometrical relationship between the line segment AB and $l$ is the distance from B to A is the least.

Taking the coordinates of A as $(a, 2a+1)$, find an equation relating $a$ and $b$ for which AB is the least.

Deduce that when AB is the least, $(AB)^2 = m (2b - 2\sqrt{b} +1)^2$ where $m$ is a constant to be found. Hence or otherwise, find the coordinates of the point on C that is nearest on $l$.

### June Revision Exercise 3 Q7

(a)

$\int x(\mathrm{ln}x)^2 ~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \int \frac{x^2}{2} 2(\mathrm{ln}x)\frac{1}{x} ~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \int x \mathrm{ln}x ~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \frac{x^2}{2} \mathrm{ln}x + \int \frac{x^2}{2}\frac{1}{x}~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \frac{x^2}{2} \mathrm{ln}x + \frac{x^2}{4} + C$

(b)

$\int \frac{1}{1-\mathrm{cos}2x} ~dx$

$= \int \frac{1}{2\mathrm{sin}^2x} ~dx$

$= \frac{2}{2} \int \mathrm{cosec}^2x~dx$

$= -\frac{1}{2} \mathrm{cot}x + C$

(c)
(i)

(ii)

$\int_{\mathrm{ln}2}^{\mathrm{ln}3} x ~dy$

$=\int_2^3 t^2 \frac{1}{t} ~dt$

$=\Big| \frac{t^2}{2} \Big|_2^3$

$= 2.5 \text{units}^2$

(iii)
$\pi \int_4^9 y^2 ~dx$

$=\pi \int_2^3 (\mathrm{ln}t)^2(2t) ~dt$

$=2\pi \int_2^3 t(\mathrm{ln}t)^2 ~dt$

$= 13.6$

Back to June Revision Exercise 3

### June Revision Exercise 3 Q6

(a)(i)

Consider $y=\sqrt{a^2 - x^2} \Rightarrow y^2+x^2=a^2$ is a circle with centre Origin and radius $a$.

Thus, required area is the area of a quadrant with radius $a$

$\text{Area} = \frac{1}{4} \pi a^2$

(ii)

$\int_0^a \frac{x^2}{\sqrt{a^2-x^2}} ~dx$

$= \int_0^2 x\frac{x}{\sqrt{a^2-x^2}} ~dx$

$= - x \sqrt{a^2-x^2} \Big|_0^a - \int_0^a -\sqrt{a^2-x^2} ~dx$

$= 0 + \frac{1}{4} \pi a^2$

$= \frac{1}{4} \pi a^2$

(b)

$u=3x+5 \Rightarrow \frac{du}{dx}=3$

$= \int_{-2}^{-\frac{5}{3}} (x-2)(3x+5)^3 ~dx$

$=\int_{-1}^0 (\frac{u-5}{3}-2)u^3 \frac{1}{4} ~du$

$=\frac{1}{9} \int_{-1}^0 u^4 - 11u^3 ~du$

$=\frac{1}{9} \Big| \frac{1}{5}u^5 - \frac{11}{4}u^4 \Big|_{-1}^0$

$= \frac{59}{180}$

Back to June Revision Exercise 3

### June Revision Exercise 4 Q7

(i)

$f(x) = \frac{4}{(1-3)\sqrt{x^2+4}}$

$= 4(1-3x)^{-1} (4)^{-\frac{1}{2}} (1 + \frac{x^2}{4})^{-\frac{1}{2}}$

$= 2[1 + 3x + (3x)^2 + (3x)^3 + \ldots][1 - \frac{1}{2} \frac{x^2}{4} + \ldots]$

$= 2[1 + 3x + (-\frac{1}{8}+9)x^2 + (-\frac{3}{8}+27)x^3 + \ldots]$

$= 2 + 6x + \frac{71}{4}x^2 + \frac{213}{4}x^2 + \ldots$

$|3x| \textless 1 \text{ and } |\frac{x^2}{4}| \textless 1$

$\Rightarrow \text{Range of values} = (-\frac{1}{3}, \frac{1}{3})$

$2 \mathrm{tan} y = f(x)$

$2 \mathrm{sec}^2 y \frac{dy}{dx} = f'(x)$

$2[2\mathrm{sec}^2 y \mathrm{tan}y (\frac{dy}{dx})^2 + \mathrm{sec}^2 y \frac{d^2y}{dx^2}] = f''(x)$

When $x = 0, f(0) = 2, f'(0)=6, f''(0)=\frac{71}{2}$

$2 \mathrm{tan}y=2 \Rightarrow y= \frac{\pi}{4}$

$2(2)\frac{dy}{dx}=6 \Rightarrow \frac{dy}{dx}=\frac{3}{2}$

$2[2(2)(1)(\frac{3}{2})^2 + 2 \frac{d^2y}{dx^2}]= \frac{71}{2} \Rightarrow = \frac{35}{8}$

Hence $y = \frac{\pi}{4} + \frac{3}{2}x + \frac{35}{8}\frac{x^2}{2!}+ \ldots = \frac{\pi}{4} + \frac{3}{2}x + \frac{35}{16}x^2 + \ldots$

### June Revision Exercise 3 Q5

(a)
$\int_0^{\mathrm{ln}2} \frac{e^3x}{e^x+2} ~dx$

$= \int_3^4 \frac{(u-2)^2}{u} ~du$

$= \int_3^4 (u-4+\frac{4}{u}) ~du$

$= \frac{1}{2}u^2 - 4u + 4 \mathrm{ln}u \Big|_3^4$

$= 4 \mathrm{ln}\frac{4}{3} - \frac{1}{2}$

(b)
$\int_0^2 |e^x - 2| dx = \int_e^{e^{\sqrt{k}}} \frac{1}{x \mathrm{ln}x} ~dx$

$- \int_0^{\mathrm{ln}2} (e^x - 2) dx + \int_{\mathrm{ln}2}^2 (e^x - 2) ~dx = \int_e^{e^{\sqrt{k}}} \frac{\frac{1}{x}}{\mathrm{ln}x} ~dx$

$-(e^x - 2x) \Big|_0^{\mathrm{ln}2} + (e^x - 2x) \Big|_{\mathrm{ln}2}^2 = \mathrm{ln}\sqrt{k}$

$\mathrm{ln}k = 2 (e^2 + 4 \mathrm{ln}2 -7)$

Back to June Revision Exercise 3

### June Revision Exercise 3 Q4

(i)
$A = 2\{ \frac{1}{n} (1) + \frac{1}{n}e^{\frac{1}{4n}} + \frac{1}{n}e^{\frac{2}{4n}} + \ldots + \frac{1}{n}e^{\frac{n-1}{4n}}\}$

$= \frac{2}{n} \{e^{\frac{0}{4n}} + e^{\frac{1}{4n}} + e^{\frac{2}{4n}} + \ldots + e^{\frac{n-1}{4n}} \}$

$= \frac{2}{n}\sum_{r=0}^{n-1} e^{\frac{r}{4n}}$

(ii)
Area of R $2 \int_0^1 e^{\frac{y}{4}} ~dy = 8(e^{\frac{1}{4}}-1)$

As $n \rightarrow \infty, A \rightarrow \text{Actual Area} = 8(e^{\frac{1}{4}}-1)$

Back to June Revision Exercise 3

### June Revision Exercise 3 Q3

(a)(i)

Let $AB=y=\frac{40-x}{2}$

$h = \sqrt{y^2-(\frac{x}{2})^2}=2\sqrt{100-5x}$

$z = \frac{1}{2}xh=\frac{1}{2}x(2\sqrt{100-5x}=x\sqrt{100-5x}$

(a)(ii)

$\frac{dz}{dx}=\sqrt{100-5x} + x\frac{1}{2}(-5)(100-5x)^{-\frac{1}{2}}$

$\frac{dz}{dx}=\frac{100-\frac{15}{2}x}{\sqrt{100-5x}}$

$\frac{dz}{dx}=0 \Rightarrow x=\frac{40}{3}$

*Students are expected to prove that $x = \frac{40}{3}$ gives the maximum area.

(b)(i)
$x^3 + 2y^3 +3xy=k$

$3x^2+6y^2\frac{dy}{dx}+3x\frac{dy}{dx}+3y=0$

$\frac{dy}{dx}=\frac{-y-x^2}{2y^2+x} (b)(ii) Tangent parallel to$latex x\$-axis $\Rightarrow \frac{dy}{dx}=0$

$y=-x^2$

Sub $y=-x^2$ into $x^3 + 2y^3 +3xy=k$

$x^3 + 2(-x^2)^3 +3x(-x^2)=k$

$2x^6 + 2x^3 + k=0$

(b)(iii)
When the line $y=-1$ is a tangent to C,

$-1=-x^2$

$x =\pm x$

When $x = 1, k=-4$

When $x = -1, k=0$

Back to June Revision Exercise 3