Sec 4 A Math: Differentiation: Techniques, practice questions & worked solutions

(a)

$$\begin{aligned} \mathrm{f}'(x) &= (2\sin 2x)(2\cos 2x)\tan 2x + \sin^2 2x\left(2\sec^2 2x\right)\\ &= 4(\sin 2x \cos 2x)\tan 2x + 2\sin^2 2x\left(\sec^2 2x\right)\\ &= 4\sin^2 2x + 2\tan^2 2x \end{aligned}$$

(b)

Given $0 < x < \dfrac{\pi}{4}$ $$\sin^2 2x > 0 \text{and} \tan^2 2x > 0,$$ hence $4\sin^2 2x + 2\tan^2 2x > 0 \implies \mathrm{f}'(x) > 0$.

Since $\mathrm{f}'(x) > 0$, $\mathrm{f}$ is an increasing function.

Method 1

$$\begin{aligned} y &= \ln\!\left(\frac{e^x}{x^2+10}\right)\\ y &= \ln e^x-\ln(x^2+10)\\ y &= x-\ln(x^2+10)\\ \frac{dy}{dx} &= 1-\frac{2x}{x^2+10}\\ \frac{dy}{dx} &= \frac{x^2+10-2x}{x^2+10}\\ \frac{dy}{dx} &= \frac{(x-1)^2+10-1}{x^2+10}\\ \frac{dy}{dx} &= \frac{(x-1)^2+9}{x^2+10} \end{aligned}$$ Since $(x-1)^2\geq 0$ then $(x-1)^2+9>0$ and $x^2+10>0$, therefore $\dfrac{(x-1)^2+9}{x^2+10}>0$ and $\dfrac{dy}{dx}\neq 0$. The curve has no turning points.

Method 2

$$\begin{aligned} y &= \ln\!\left(\frac{e^x}{x^2+10}\right)\\ y &= \ln e^x-\ln(x^2+10)\\ y &= x-\ln(x^2+10)\\ \frac{dy}{dx} &= 1-\frac{2x}{x^2+10}\\ \frac{x^2+10-2x}{x^2+10} &= 0\\ x^2+10-2x &= 0\\ x &= \frac{-2(-2)\pm\sqrt{(-2)^2-4(1)(10)}}{2} \text{no solutions or } b^2-4ac<0 \end{aligned}$$ $\dfrac{dy}{dx}\neq 0$. The curve has no turning points.

(a)

$$\begin{aligned} y &= \frac{\sin 2x}{2 - \cos 2x}\\ \frac{dy}{dx} &= \frac{(2 - \cos 2x)2\cos 2x - \sin 2x(2\sin 2x)}{(2 - \cos 2x)^2}\\ &= \frac{4\cos 2x - 2\cos^2 2x - 2\sin^2 2x}{(2 - \cos 2x)^2}\\ &= \frac{4\cos 2x - 2(\cos^2 2x + \sin^2 2x)}{(2 - \cos 2x)^2}\\ &= \frac{4\cos 2x - 2}{(2 - \cos 2x)^2} \end{aligned}$$

(b)

$$\begin{aligned} \text{gradient of tangent} &= \frac{4\cos\frac{\pi}{6} - 2}{\left(2 - \cos\frac{\pi}{6}\right)^2} = 1.1386\\ \text{Angle required} &= \tan^{-1}(1.1386) = 48.7^\circ \ (\text{1 dp}) \end{aligned}$$

(a)

$$\begin{aligned} \frac{d}{dx}\left[x(x - 3)^4\right] &= (x - 3)^4 + 4x(x - 3)^3\\ &= (x - 3)^3(x - 3 + 4x)\\ &= (5x - 3)(x - 3)^3 \end{aligned}$$

(b)

$$y = \frac{x(x - 3)^4}{5x - 3}$$ $$\begin{aligned} \frac{dy}{dx} &= \frac{(5x - 3)(x - 3)^3(5x - 3) - 5x(x - 3)^4}{(5x - 3)^2}\\ &= \frac{(x - 3)^3\big[(5x - 3)^2 - 5x(x - 3)\big]}{(5x - 3)^2}\\ &= \frac{(x - 3)^3(25x^2 - 30x + 9 - 5x^2 + 15x)}{(5x - 3)^2}\\ &= \frac{(x - 3)^3(20x^2 - 15x + 9)}{(5x - 3)^2} \end{aligned}$$ When $x = 1$, $$\begin{aligned} \frac{dy}{dt} &= \frac{(x - 3)^3(20x^2 - 15x + 9)}{(5x - 3)^2} \times \frac{dx}{dt}\\ -70 &= \frac{(1 - 3)^3(20 - 15 + 9)}{(5 - 3)^2} \times \frac{dx}{dt}\\ -70 &= -28 \times \frac{dx}{dt}\\ \end{aligned}$$ $$\mathbf{2.5\text{ units/s}}$$

(a)

$$\begin{aligned} \mathrm{f}'(x) &= \frac{(x^2 + 3) - 2x(x + a)}{(x^2 + 3)^2}\\ &= \frac{-x^2 - 2ax + 3}{(x^2 + 3)^2} \end{aligned}$$

(b)

$$\frac{-x^2 - 2ax + 3}{(x^2 + 3)^2} > 0, \ \text{for } b < x < 1$$ Since $(x^2 + 3)^2 > 0$, $$\begin{aligned} -x^2 - 2ax + 3 &> 0\\ x^2 + 2ax - 3 &< 0 \end{aligned}$$ Given that $b < x < 1$ $$\begin{aligned} (x - b)(x - 1) &< 0\\ x^2 - bx - x + b &< 0 \end{aligned}$$ By comparing: $$\begin{aligned} b &= -3\\ 2a &= -b - 1\\ 2a &= 3 - 1\\ a &= 1 \end{aligned}$$

(a)

$$q=-2$$

(b)

$$\begin{aligned} y &= \frac{2x-1}{\sqrt{x+2}}\\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{2\sqrt{x+2}-(2x-1)\cdot\frac{1}{2}(x+2)^{-\frac{1}{2}}(1)}{x+2}\\ &= \frac{2\sqrt{x+2}-\dfrac{2x-1}{2\sqrt{x+2}}}{x+2}\\ &= \frac{4(x+2)-(2x-1)}{2\sqrt{(x+2)^3}}\\ &= \frac{2x+9}{2\sqrt{(x+2)^3}} \end{aligned}$$

(c)

For $x>-2$, $$\sqrt{(x+2)^3}>0, 2\sqrt{(x+2)^3}>0\ \text{ and }\ 2x+9>0$$ $$\Rightarrow\ \frac{\mathrm{d}y}{\mathrm{d}x}>0$$ $y=\dfrac{2x-1}{\sqrt{x+2}}$ is an increasing function.

$$\begin{aligned} \frac{dy}{dx} &= \frac{(3x^2)(e^{2x})(2) - (e^{2x})(6x)}{(3x^2)^2}\\ &= \frac{6xe^{2x}(x - 1)}{9x^4}\\ &= \frac{2e^{2x}(x - 1)}{3x^3}\\ 0 &= \frac{2e^{2x}(x - 1)}{3x^3} \end{aligned}$$ $$\begin{aligned} 2e^{2x} &> 0,\ x \neq 0\\ x - 1 &= 0\\ x &= 1\\ y &= \frac{e^2}{3} \end{aligned}$$ $$\text{Stationary point} = \left(1,\ \frac{e^2}{3}\right)$$

(a)

$$\begin{aligned} y &= \frac{m - 2x}{x^2 - 3}\\ \frac{dy}{dx} &= \frac{\left(x^2 - 3\right)(-2) - (m - 2x)(2x)}{\left(x^2 - 3\right)^2}\\ &= \frac{-2x^2 + 6 - 2mx + 4x^2}{\left(x^2 - 3\right)^2}\\ &= \frac{2x^2 + 6 - 2mx}{\left(x^2 - 3\right)^2} \end{aligned}$$

(b)

$$\begin{aligned} \text{Since } \frac{2x^2 + 6 - 2mx}{\left(x^2 - 3\right)^2} > 0 \ &\text{and} \ \left(x^2 - 3\right)^2 > 0,\\ \therefore \ 2x^2 - 2mx + 6 &> 0\\ 2x^2 - 2mx + 6 &= 2\left(x^2 - mx + 3\right)\\ &= 2\left[\left(x - \frac{m}{2}\right)^2 - \left(\frac{m}{2}\right)^2 + 3\right]\\ &= 2\left(x - \frac{m}{2}\right)^2 - \frac{m^2}{2} + 6 \end{aligned}$$ For $2x^2 - 2mx + 6 > 0$, $$\begin{aligned} -\frac{m^2}{2} + 6 &> 0\\ \frac{m^2}{2} &< 6\\ m^2 &< 12\\ m^2 - 12 &< 0\\ \left(m + \sqrt{12}\right)\left(m - \sqrt{12}\right) &< 0\\ -2\sqrt{3} < m &< 2\sqrt{3} \end{aligned}$$

Alternative Solution

$$\begin{aligned} \text{Since } \frac{2x^2 + 6 - 2mx}{\left(x^2 - 3\right)^2} > 0 \ &\text{and} \ \left(x^2 - 3\right)^2 > 0,\\ \therefore \ 2x^2 - 2mx + 6 &> 0 \end{aligned}$$ $$\begin{aligned} 2x^2 - 2mx + 6 &= 0\\ \text{discriminant} &< 0\\ (-2m)^2 - 4(2)(6) &< 0\\ 4m^2 - 48 &< 0\\ m^2 - 12 &< 0\\ \left(m + \sqrt{12}\right)\left(m - \sqrt{12}\right) &< 0\\ -2\sqrt{3} < m &< 2\sqrt{3} \end{aligned}$$

(a)

$$\begin{aligned} \frac{d}{dx}\bigl[\ln(\tan 3x)\bigr] &= \frac{1}{\tan 3x}\left(\sec^2 3x\right)(3)\\ &= \frac{3\cos 3x}{\sin 3x\cos^2 3x}\\ &= \frac{3}{\sin 3x\cos 3x}\\ &= \frac{3(2)}{2\sin 3x\cos 3x}\\ &= \frac{6}{\sin 6x}\\ k &= 6 \end{aligned}$$

(b)

$$\begin{aligned} \frac{d}{dx}\left(\ln\sqrt{\frac{7x}{x^2 - 1}}\right) &= \frac{d}{dx}\left(\frac{1}{2}\ln\frac{7x}{x^2 - 1}\right)\\ &= \frac{d}{dx}\left(\frac{1}{2}\Bigl[\ln 7x - \ln\left(x^2 - 1\right)\Bigr]\right)\\ &= \frac{1}{2}\left(\frac{7}{7x}\right) - \frac{1}{2}\left(\frac{2x}{x^2 - 1}\right)\\ &= \frac{1}{2}\left(\frac{x^2 - 1 - 2x^2}{x\left(x^2 - 1\right)}\right)\\ &= -\frac{x^2 + 1}{2x\left(x^2 - 1\right)} \end{aligned}$$

(a)

$$\begin{aligned} \frac{d}{dx}(\cot x) &= \frac{d}{dx}\left(\frac{\cos x}{\sin x}\right)\\ &= \frac{\sin x(-\sin x) - \cos x(\cos x)}{\sin^2 x}\\ &= \frac{-\sin^2 x - \cos^2 x}{\sin^2 x}\\ &= \frac{-1}{\sin^2 x} \text{or} -1 - \cot^2 x\\ &= -\cosec^2 x \end{aligned}$$

Or

$$\begin{aligned} \frac{d}{dx}(\cot x) &= \frac{d}{dx}\left(\frac{1}{\tan x}\right) = \frac{d}{dx}(\tan x)^{-1}\\ &= -1(\tan x)^{-2}(\sec^2 x)\\ &= -\frac{1}{\tan^2 x\cos^2 x}\\ &= \frac{-1}{\sin^2 x}\\ &= -\cosec^2 x \end{aligned}$$

(b)

$$\begin{aligned} \frac{dy}{dx} &= \mathrm{e}^{-ax}\left(-\cosec^2 x\right) - a\mathrm{e}^{-ax}\cot x\\ &= -\mathrm{e}^{-ax}\left(\cosec^2 x + a\cot x\right) \end{aligned}$$ The tangent is parallel to $x$-axis means $\dfrac{dy}{dx} = 0$. $$-\mathrm{e}^{-ax}\left(\cosec^2 x + a\cot x\right) = 0$$ Since $\mathrm{e}^{-ax} > 0$ for all values of $x$, $$\begin{aligned} -\left(\cosec^2 x + a\cot x\right) &= 0\\ \cosec^2 x + a\cot x &= 0\\ \cot^2 x + 1 + a\cot x &= 0 \end{aligned}$$ Since there is only 1 point, the above equation only has 1 solution, i.e. discriminant $= 0$. $$\begin{aligned} a^2 - 4(1)(1) &= 0\\ a &= \pm 2\\ a &= 2 \ (a > 0)\\ \cot^2 x + 2\cot x + 1 &= 0\\ (\cot x + 1)^2 &= 0\\ \cot x &= -1\\ \tan x &= -1\\ x &= -\frac{\pi}{4} \end{aligned}$$