Limits, derivatives, differentiation rules, tangents and normals – IB Maths exam questions with mark scheme

1CoreNon-calculator4 marks

Let $f(x)=4x^3-\dfrac{6}{x}+\sqrt{x}$ for $x>0$.

(a) Find $f'(x)$. [3]

(b) Find the value of $f'(1)$. [1]

$12x^2$ (derivative of $4x^3$)A1
$+\dfrac{6}{x^2}$ (from $-6x^{-1}$; accept $6x^{-2}$)A1
$+\dfrac{1}{2\sqrt{x}}$ (accept $\tfrac12x^{-1/2}$)A1
$f'(1)=12+6+\tfrac12=\dfrac{37}{2}$ (accept 18.5)A1
Answer: (a) $f'(x)=12x^2+\dfrac{6}{x^2}+\dfrac{1}{2\sqrt x}$; (b) $\dfrac{37}{2}$
Examiner tip: Rewrite $6/x$ and $\sqrt{x}$ as powers first; the sign on the $6/x^2$ term is the usual slip (two negatives make a positive).

2StandardNon-calculator6 marks

The curve $C$ has equation $y=x^3-3x^2+2$. The point $P$ on $C$ has $x$-coordinate 3.

(a) Find the equation of the tangent to $C$ at $P$, giving your answer in the form $y=mx+c$. [4]

(b) Find the equation of the normal to $C$ at $P$, giving your answer in the form $ax+by+d=0$ where $a,b,d\in\mathbb{Z}$. [2]

$y=2$ at $x=3$, so $P=(3,2)$B1
Attempts $\dfrac{dy}{dx}=3x^2-6x$ (at least one term correct)M1
Gradient at $P$ is $27-18=9$A1
$y-2=9(x-3)$, so $y=9x-25$A1
Normal gradient $-\dfrac19$ (negative reciprocal) used with $P$: $y-2=-\dfrac19(x-3)$M1
$x+9y-21=0$ (or any integer multiple)A1
Answer: (a) $y=9x-25$; (b) $x+9y-21=0$
Examiner tip: Find the $y$-coordinate of $P$ before anything else, and clear the fraction in (b) because the question demands integer coefficients.

3StandardNon-calculator5 marks

Differentiate with respect to $x$:

(a) $y=x^2e^{3x}$, giving your answer in a fully factorised form; [2]

(b) $y=\dfrac{\sin 2x}{x}$. [3]

Product rule: $2x\,e^{3x}+x^2\cdot 3e^{3x}$M1
$x\,e^{3x}(2+3x)$A1
Quotient rule form $\dfrac{x\cdot(\ldots)-\sin 2x\cdot 1}{x^2}$M1
Numerator correct: $x\cdot2\cos2x-\sin2x\cdot1$ (derivative of $\sin2x$ is $2\cos2x$)A1
$\dfrac{dy}{dx}=\dfrac{2x\cos 2x-\sin 2x}{x^2}$A1
Answer: (a) $xe^{3x}(3x+2)$; (b) $\dfrac{2x\cos2x-\sin2x}{x^2}$
Examiner tip: The chain rule factor 2 from $\sin 2x$ and factor 3 from $e^{3x}$ are the marks most often dropped.

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4HardNon-calculator5 marks

The line $y=2x+k$, where $k\in\mathbb{R}$, is a tangent to the curve $y=x^3-x$. Find all possible values of $k$.

$\dfrac{dy}{dx}=3x^2-1$M1
Sets gradient equal to the line's gradient: $3x^2-1=2$M1
$x=1$ or $x=-1$A1
Uses the point on the curve ($y=0$ at both $x=\pm1$) in $y=2x+k$M1
$k=-2$ and $k=2$ (both required)A1
Answer: $k=-2$ or $k=2$
Examiner tip: There are two tangent points; stopping after $x=1$ loses the final mark.

5CoreCalculator6 marks

Let $f(x)=x^3-2x$. The point $P(2,\,4)$ lies on the graph of $f$. The point $Q$ also lies on the graph of $f$ and has $x$-coordinate $2+h$, where $h\neq0$.

(a) Copy and complete the table, giving the gradient of the chord $PQ$ for each value of $h$. [2]

$h$	$0.5$	$0.1$	$0.01$
Gradient of $PQ$	$13.25$

(b) Show that the gradient of $PQ$ is $10+6h+h^2$. [2]

(c) Hence write down the value of $\displaystyle\lim_{h\to0}\left(10+6h+h^2\right)$ and state what this value represents for the graph of $f$. [2]

$10.61$ for $h=0.1$A1
$10.0601$ for $h=0.01$ (accept $10.1$ (3 s.f.) or $10.06$)A1
Substitutes $x=2+h$ and forms $\dfrac{(2+h)^3-2(2+h)-4}{h}$ (allow $f(2+h)-f(2)$ over $(2+h)-2$)M1
Expands correctly to $\dfrac{10h+6h^2+h^3}{h}$ and divides by $h$ to reach $10+6h+h^2$ (AG; no errors seen)A1
Limit $=10$A1
It is the gradient of the tangent to the graph of $f$ at $P$ (accept $f'(2)$, instantaneous rate of change of $f$ at $x=2$; do not accept just 'the gradient of the chord')R1
Answer: (a) $10.61$, $10.0601$; (c) $10$, the gradient of the tangent (the derivative $f'(2)$) at $P$
Examiner tip: In (b) the answer is given, so every line of the expansion must be shown; the mark is lost if $(2+h)^3$ appears already expanded with no working.

6StandardNon-calculator6 marks

The functions $f$ and $g$ are differentiable for all $x\in\mathbb{R}$, with $f(x)>0$. The table shows some values of $f$, $g$ and their derivatives.

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
$1$	$2$	$6$	$3$	$-2$
$3$	$4$	$5$	$1$	$4$

(a) Let $h(x)=f\big(g(x)\big)$. Find $h'(1)$. [2]

(b) Let $p(x)=\dfrac{g(x)}{f(x)}$. Find $p'(1)$. [2]

Uses chain rule: $h'(1)=f'\big(g(1)\big)\times g'(1)=f'(3)\times(-2)$M1
$h'(1)=-10$A1
Uses quotient rule with correct values: $\dfrac{(-2)(2)-(3)(6)}{2^2}$ (allow numerator terms reversed in sign only if the formula is quoted correctly first)M1
$p'(1)=-\dfrac{22}{4}=-\dfrac{11}{2}$ (accept $-5.5$)A1
$q'(x)=\dfrac{f'(x)}{f(x)}$ (seen or implied)M1
$q'(3)=\dfrac54$ (accept $1.25$)A1
Answer: (a) $-10$; (b) $-\dfrac{11}{2}$; (c) $\dfrac54$
Examiner tip: In (a) the outer derivative is evaluated at $g(1)=3$, not at $1$; using $f'(1)=6$ to get $-12$ scores M0.

7StandardNon-calculator7 marks

The diagram shows the graph of the derivative $y=f'(x)$ for $-2.5\le x\le3.5$. The graph of $f'$ crosses the $x$-axis at $(-2,0)$, $(1,0)$ and $(3,0)$, and the $y$-axis at $(0,3)$.

(a) Write down the $x$-coordinate of the local maximum point on the graph of $f$. Justify your answer. [2]

(b) Write down the intervals for which $f$ is decreasing. [2]

(d) Find the gradient of the normal to the graph of $f$ at the point where $x=0$. [1]

$x=1$A1
$f'(x)$ changes from positive to negative at $x=1$ (accept graph of $f'$ goes from above to below the $x$-axis); R0 for '$f'(1)=0$' aloneR1
$-2.5\le x\le-2$ (accept strict inequalities, or $-2.5\le x<-2$)A1
$1\le x\le3$ (accept $1A1
$f'(x)>0$ for $0R1
$f(0)$ is less than $f(1)$ (R1 must be awarded for this mark)A1
$f'(0)=3$ read from graph, so normal gradient $=-\dfrac13$A1
Answer: (a) $x=1$, since $f'$ changes from $+$ to $-$; (b) $-2.5\le x\le-2$ and $1\le x\le3$; (c) $f(0)0$ on $0
Examiner tip: This is the graph of $f'$, not $f$: the turning points of $f$ are where this graph crosses the $x$-axis, not where it has its own turning points.

8HardNon-calculator7 marks

Consider the curve $y=\sqrt{2x+1}$, $x\ge-\tfrac12$. The point $A(4,\,3)$ lies on the curve.

(a) Find the equation of the tangent to the curve at $A$. [3]

(b) Find the equation of the normal to the curve at $A$. [2]

The tangent meets the $x$-axis at $B$ and the normal meets the $x$-axis at $C$.

Attempts chain rule: $\dfrac{dy}{dx}=\dfrac12(2x+1)^{-1/2}\times2$ (allow missing $\times2$ for M1)M1
$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{2x+1}}$, gradient at $A$ $=\dfrac13$A1
$y-3=\dfrac13(x-4)$ or $y=\dfrac13x+\dfrac53$ (accept any correct form)A1
Normal gradient $-3$ used with $A$: $y-3=-3(x-4)$ (ft their tangent gradient)M1
$y=-3x+15$ (accept any correct form)A1
$B(-5,0)$ and $C(5,0)$ (both)A1
Area $=\dfrac12\times10\times3=15$A1
Answer: (a) $y=\dfrac13x+\dfrac53$; (b) $y=-3x+15$; (c) $15$
Examiner tip: Forgetting the inner derivative $2$ gives a gradient of $\tfrac16$ and costs every later accuracy mark; the height of the triangle is the $y$-coordinate of $A$.

9HardNon-calculator7 marks

Let $f(x)=e^{2x}$. The tangent to the graph of $f$ at the point where $x=a$ passes through the point $(1,\,0)$.

(a) Show that $a=\dfrac32$. [4]

(b) Find the equation of this tangent, giving your answer in the form $y=mx+c$ where $m$ and $c$ are exact. [1]

The normal to the graph of $f$ at $x=a$ meets the $x$-axis at the point $D$.

$f'(x)=2e^{2x}$A1
Equation of tangent at $x=a$: $y-e^{2a}=2e^{2a}(x-a)$M1
Substitutes $(1,0)$: $-e^{2a}=2e^{2a}(1-a)$M1
Divides by $e^{2a}\ (\neq0)$: $-1=2-2a$, so $a=\dfrac32$ (AG; must see the division or cancellation justified)A1
$y=2e^3x-2e^3$A1
Normal: $y-e^3=-\dfrac{1}{2e^3}\left(x-\dfrac32\right)$ and sets $y=0$M1
$x=\dfrac32+2e^6$A1
Answer: (a) shown; (b) $y=2e^3x-2e^3$; (c) $\dfrac32+2e^6$
Examiner tip: Keep $a$ general in the tangent equation and substitute $(1,0)$ before solving; plugging $a=\tfrac32$ in to check it works is not a 'show that' and scores at most the A1 for $f'$.

10StretchNon-calculatorHL only9 marks

Let $f(x)=\dfrac{1}{2x+1}$, $x\neq-\dfrac12$.

(a) Use differentiation from first principles to show that $f'(x)=-\dfrac{2}{(2x+1)^2}$. [4]

(b) Find the equation of the normal to the graph of $f$ at the point where $x=\dfrac12$. [3]

Correct use of the definition: $f'(x)=\displaystyle\lim_{h\to0}\dfrac{\frac{1}{2(x+h)+1}-\frac{1}{2x+1}}{h}$M1
Combines into a single fraction: $\dfrac{(2x+1)-(2x+2h+1)}{h(2x+2h+1)(2x+1)}$M1
Simplifies and cancels $h$: $\dfrac{-2}{(2x+2h+1)(2x+1)}$A1
Lets $h\to0$ to reach $-\dfrac{2}{(2x+1)^2}$ (AG; the limit notation must be present throughout or clearly applied at the last step)A1
$f\left(\tfrac12\right)=\dfrac12$ and $f'\left(\tfrac12\right)=-\dfrac12$A1
Normal gradient $2$ used: $y-\dfrac12=2\left(x-\dfrac12\right)$M1
$y=2x-\dfrac12$ (accept $4x-2y-1=0$)A1
Sets $-\dfrac{2}{(2x+1)^2}=-\dfrac29$, giving $(2x+1)^2=9$M1
$x=1$ or $x=-2$ (both; A0 if only $x=1$)A1
Answer: (a) shown; (b) $y=2x-\dfrac12$; (c) $x=1,\ x=-2$
Examiner tip: Students lose the final mark in (a) by letting $h=0$ before the $h$ in the denominator has been cancelled, or by dropping the $\lim$ notation.

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