Moderate -0.3 This is a straightforward multi-part differentiation question testing standard techniques: quotient rule with trig/power functions, chain rule with trig identities, and implicit differentiation with inverse trig. All parts are routine applications of C3 techniques with no novel problem-solving required, making it slightly easier than average.
5. (a) Differentiate
$$\frac { \cos 2 x } { \sqrt { x } }$$
with respect to \(x\).
(b) Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sec ^ { 2 } 3 x \right)\) can be written in the form
$$\mu \left( \tan 3 x + \tan ^ { 3 } 3 x \right)$$
where \(\mu\) is a constant.
(c) Given \(x = 2 \sin \left( \frac { y } { 3 } \right)\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\), simplifying your answer.
Rule must be correct if quoted; if not quoted accept form \(\frac{\sqrt{x}\times \pm A\sin 2x - \cos 2x \times Bx^{-\frac{1}{2}}}{(\sqrt{x})^2 \text{ or } x^{\frac{1}{4}}}\)
Alt (a) using product rule:
Answer
Marks
Guidance
Applies \(vu'+uv'\) to \(x^{-\frac{1}{2}}\cos 2x\)
M1
Accept form \(\pm Ax^{-\frac{1}{2}}\sin 2x - Bx^{-\frac{3}{2}}\cos 2x\)
Inverting \(\frac{dx}{dy}\) and using \(\sin^2\frac{y}{3}+\cos^2\frac{y}{3}=1\) to get \(\frac{dy}{dx}\) in terms of \(x\) only; dependent on first M1
\(\frac{dy}{dx}=\frac{3}{\sqrt{4-x^2}}\)
A1 (cao)
Must be in simplest form. Do not accept \(\frac{3}{2\sqrt{1-\frac{1}{4}x^2}}\) or \(\frac{1}{\frac{1}{3}\sqrt{4-x^2}}\)
Alt 5(c): \(y=3\arcsin\!\left(\frac{x}{2}\right)\Rightarrow \frac{dy}{dx}=\frac{3}{\sqrt{1-\left(\frac{x}{2}\right)^2}}\times\frac{1}{2}\)
M1dM1A1
M1: rearrange to \(y=A\arcsin Bx\), differentiate to \(\frac{dy}{dx}=\frac{A}{\sqrt{1-Bx^2}}\); dM1: form of rhs must be correct \(\frac{dy}{dx}=\frac{C}{\sqrt{1-\left(\frac{x}{2}\right)^2}}\); A1: correct unsimplified answer
\(\frac{dy}{dx}=\frac{3}{\sqrt{4-x^2}}\)
A1
## Question 5:
### Part (a):
| $\frac{d}{dx}(\cos 2x)=-2\sin 2x$ | B1 | Award for sight of this; could be seen in their differential |
| Applies $\frac{vu'-uv'}{v^2}$ to $\frac{\cos 2x}{\sqrt{x}}$: $\displaystyle\frac{\sqrt{x}\times -2\sin 2x - \cos 2x \times \frac{1}{2}x^{-\frac{1}{2}}}{(\sqrt{x})^2}$ | M1A1 | Rule must be correct if quoted; if not quoted accept form $\frac{\sqrt{x}\times \pm A\sin 2x - \cos 2x \times Bx^{-\frac{1}{2}}}{(\sqrt{x})^2 \text{ or } x^{\frac{1}{4}}}$ |
**Alt (a) using product rule:**
| Applies $vu'+uv'$ to $x^{-\frac{1}{2}}\cos 2x$ | M1 | Accept form $\pm Ax^{-\frac{1}{2}}\sin 2x - Bx^{-\frac{3}{2}}\cos 2x$ |
| $-2x^{-\frac{1}{2}}\sin 2x - \frac{1}{2}x^{-\frac{3}{2}}\cos 2x$ | A1 | Correct answer; need not be simplified |
### Part (b):
| $\frac{d}{dx}(\sec^2 3x)=2\sec 3x \times 3\sec 3x\tan 3x\ (=6\sec^2 3x\tan 3x)$ | M1 | Correct application of chain rule on $\sec^2 3x$; sight of $C\sec 3x\sec 3x\tan 3x$ sufficient |
| $=6(1+\tan^2 3x)\tan 3x$ | dM1 | Replacing $\sec^2 3x=1+\tan^2 3x$; dependent on first M |
| $=6(\tan 3x+\tan^3 3x)$ | A1 | No need to write $\mu=6$ |
**Alt (b) product rule:** Writes $\sec^2 3x$ as $\sec 3x \times \sec 3x$, uses product rule with $u'=A\sec 3x\tan 3x$, $v'=B\sec 3x\tan 3x$ | M1 | Form $A\sec 3x\tan 3x\sec 3x + B\sec 3x\tan 3x\sec 3x$ |
| Replace $\sec^2 3x$ with $1+\tan^2 3x$ to get expression in $\tan 3x$ only | dM1 | Dependent on first M |
**Alt (b) using** $\sec 3x=\frac{1}{\cos 3x}$:
| Writes $\sec^2 3x$ as $(\cos 3x)^{-2}$, differentiates to $A(\cos 3x)^{-3}\sin 3x$ | M1 | |
| Uses $\frac{\sin 3x}{\cos 3x}=\tan 3x$, $\frac{1}{\cos^2 3x}=\sec^2 3x$, $\sec^2 3x=1+\tan^2 3x$ | dM1 | Dependent on first M |
| $6(\tan 3x+\tan^3 3x)$ | A1 | |
### Part (c):
| $x=2\sin\!\left(\frac{y}{3}\right)\Rightarrow \frac{dx}{dy}=\frac{2}{3}\cos\!\left(\frac{y}{3}\right)$ | M1A1 | Award for knowing $\sin\!\left(\frac{y}{3}\right)$ differentiates to $\cos\!\left(\frac{y}{3}\right)$; lhs need not be correct |
| $\frac{dy}{dx}=\frac{1}{\frac{2}{3}\cos\!\left(\frac{y}{3}\right)}=\frac{1}{\frac{2}{3}\sqrt{1-\sin^2\!\left(\frac{y}{3}\right)}}=\frac{1}{\frac{2}{3}\sqrt{1-\left(\frac{x}{2}\right)^2}}$ | dM1 | Inverting $\frac{dx}{dy}$ and using $\sin^2\frac{y}{3}+\cos^2\frac{y}{3}=1$ to get $\frac{dy}{dx}$ in terms of $x$ only; dependent on first M1 |
| $\frac{dy}{dx}=\frac{3}{\sqrt{4-x^2}}$ | A1 (cao) | Must be in simplest form. Do not accept $\frac{3}{2\sqrt{1-\frac{1}{4}x^2}}$ or $\frac{1}{\frac{1}{3}\sqrt{4-x^2}}$ |
**Alt 5(c):** $y=3\arcsin\!\left(\frac{x}{2}\right)\Rightarrow \frac{dy}{dx}=\frac{3}{\sqrt{1-\left(\frac{x}{2}\right)^2}}\times\frac{1}{2}$ | M1dM1A1 | M1: rearrange to $y=A\arcsin Bx$, differentiate to $\frac{dy}{dx}=\frac{A}{\sqrt{1-Bx^2}}$; dM1: form of rhs must be correct $\frac{dy}{dx}=\frac{C}{\sqrt{1-\left(\frac{x}{2}\right)^2}}$; A1: correct unsimplified answer |
| $\frac{dy}{dx}=\frac{3}{\sqrt{4-x^2}}$ | A1 | |
5. (a) Differentiate
$$\frac { \cos 2 x } { \sqrt { x } }$$
with respect to $x$.\\
(b) Show that $\frac { \mathrm { d } } { \mathrm { d } x } \left( \sec ^ { 2 } 3 x \right)$ can be written in the form
$$\mu \left( \tan 3 x + \tan ^ { 3 } 3 x \right)$$
where $\mu$ is a constant.\\
(c) Given $x = 2 \sin \left( \frac { y } { 3 } \right)$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$, simplifying your answer.
\hfill \mbox{\textit{Edexcel C3 2013 Q5 [10]}}