Pre-U Pre-U 9794/2 2017 June — Question 5 7 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
Year2017
SessionJune
Marks7
TopicTangents, normals and gradients
TypeIncreasing/decreasing intervals
DifficultyModerate -0.3 Part (i) requires quotient rule differentiation with a composite function in the denominator—a standard technique. Part (ii) is straightforward: show the derivative is positive for all x, which follows immediately from the form of the derivative. This is a routine calculus exercise with no novel insight required, making it slightly easier than average.
Spec1.07o Increasing/decreasing: functions using sign of dy/dx1.07q Product and quotient rules: differentiation
5
  1. Differentiate \(\frac { x } { \sqrt { 1 + x ^ { 2 } } }\) with respect to \(x\).
  2. Hence show that \(\frac { x } { \sqrt { 1 + x ^ { 2 } } }\) is increasing for all \(x\).
Question 5: Differentiate \(\frac{x}{\sqrt{1+x^2}}\)
(i)
Method 1 (Quotient rule):
- \(\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{1+x^2} = \frac{x}{\sqrt{1+x^2}}\) M1 Attempt use of chain rule to obtain \(kx(1+x^2)^{-\frac{1}{2}}\)
- A1 Obtain correct derivative, soi
- \(\frac{\mathrm{d}}{\mathrm{d}x}\frac{x}{\sqrt{1+x^2}} = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2}\) M1 Attempt use of quotient rule (allow \(uv' - u'v\) in numerator)
- A1 Obtain correct numerator or denominator – must now be from correct rule
- A1 Obtain correct derivative aef
Method 2 (Product rule):
- \(\frac{\mathrm{d}}{\mathrm{d}x}(1+x^2)^{-\frac{1}{2}} = -x(1+x^2)^{-\frac{3}{2}}\) M1 Attempt use of chain rule to obtain \(kx(1+x^2)^{-\frac{3}{2}}\)
- A1 Obtain correct derivative, soi
- M1 Attempt use of product rule
- \(\frac{\mathrm{d}}{\mathrm{d}x}x(1+x^2)^{-\frac{1}{2}} = (1+x^2)^{-\frac{1}{2}} - x^2(1+x^2)^{-\frac{3}{2}}\) A1 Obtain one correct term – from correct rule
- A1 Obtain correct derivative aef
(ii)
- \(\frac{\mathrm{d}}{\mathrm{d}x}\frac{x}{\sqrt{1+x^2}} = \frac{1}{(1+x^2)^{\frac{3}{2}}}\) B1\* Simplify to correct useable form (may be seen in part (i))
- \(1 + x^2 > 0\) so it is increasing B1d\* Conclude appropriately – must refer to both positive gradient (could be algebraic) and increasing
Total: 7 marks
**Question 5: Differentiate $\frac{x}{\sqrt{1+x^2}}$**

**(i)**

**Method 1 (Quotient rule):**
- $\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{1+x^2} = \frac{x}{\sqrt{1+x^2}}$ **M1** Attempt use of chain rule to obtain $kx(1+x^2)^{-\frac{1}{2}}$
- **A1** Obtain correct derivative, soi
- $\frac{\mathrm{d}}{\mathrm{d}x}\frac{x}{\sqrt{1+x^2}} = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2}$ **M1** Attempt use of quotient rule (allow $uv' - u'v$ in numerator)
- **A1** Obtain correct numerator or denominator – must now be from correct rule
- **A1** Obtain correct derivative aef

**Method 2 (Product rule):**
- $\frac{\mathrm{d}}{\mathrm{d}x}(1+x^2)^{-\frac{1}{2}} = -x(1+x^2)^{-\frac{3}{2}}$ **M1** Attempt use of chain rule to obtain $kx(1+x^2)^{-\frac{3}{2}}$
- **A1** Obtain correct derivative, soi
- **M1** Attempt use of product rule
- $\frac{\mathrm{d}}{\mathrm{d}x}x(1+x^2)^{-\frac{1}{2}} = (1+x^2)^{-\frac{1}{2}} - x^2(1+x^2)^{-\frac{3}{2}}$ **A1** Obtain one correct term – from correct rule
- **A1** Obtain correct derivative aef

**(ii)**
- $\frac{\mathrm{d}}{\mathrm{d}x}\frac{x}{\sqrt{1+x^2}} = \frac{1}{(1+x^2)^{\frac{3}{2}}}$ **B1\*** Simplify to correct useable form (may be seen in part (i))
- $1 + x^2 > 0$ so it is increasing **B1d\*** Conclude appropriately – must refer to both positive gradient (could be algebraic) and increasing

**Total: 7 marks**
5 (i) Differentiate $\frac { x } { \sqrt { 1 + x ^ { 2 } } }$ with respect to $x$.\\
(ii) Hence show that $\frac { x } { \sqrt { 1 + x ^ { 2 } } }$ is increasing for all $x$.

\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2017 Q5 [7]}}
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