Further Differentiation (DP IB Analysis & Approaches (AA): HL): Exam Questions

Let $f\left(x\right)=2x^3$.

By differentiating from first principles, show that $f\text{'}\left(x\right)=6x^2$.

Let $f\left(x\right)=\cos \sqrt{2}x$.

Find the positive solution to the equation
$f^{\text{'}}\left(x\right)=f^{\text{'}\text{'}\text{'}}\left(x\right)-\frac{3\sqrt{2}}{2}$   that is closest to zero.

(i) Show that $f^{\left(4\right)}\left(x\right)=kf\left(x\right)$,  where $k$ is a constant to be determined.

(ii) Write down the value of $f^{\left(10\right)}\left(0\right)$.

Given that $f\left(x\right)=\arctan \left(\frac{1}{x}\right),$  find $f\text{'}\left(x\right)$.

For the function g defined by g$\left(x\right)=4^{-x}+2{\log }_{4}x$,  show that

$g^{\text{'}}\left(x\right)=\frac{1}{\left(\ln 2\right)x}-\left(\ln 2\right)2^{-2x+1}$

Find the derivative of the function $h\left(x\right)=\mathrm{cosec}\left(3x^2\right)$.

For the curve defined by $y=\cot x^2$,  show that

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=8x^2\left(y^3+y\right)-2y^2-2$

For the curve defined by $y=\arcsin x$,  show that

$y^{\text{'}\text{'}}=\frac{x}{\left(1-x^2\right)\sqrt{1-x^2}}$

Consider the function $f$ defined by $f\left(x\right)=\tan \left(\frac{\mathrm{\pi }}{2}-x\right)+\frac{4}{3}x,0<x<\mathrm{\pi }$. 

The following diagram shows the graph of the curve $y=f\left(x\right):$

The points marked A and B are the turning points of the graph. 

Find the coordinates of points A and B.

Find the equation of the normal to the graph at the point where the $x$-coordinate is equal to $\frac{\mathrm{\pi }}{4}$.

For each of the following, find $\frac{dy}{dx}$ by differentiating implicitly with respect to $x$.

$\frac{x^2}{16}+\frac{y^2}{25}=1$

$e^{x^2}+x=y^3-y$

$\frac{{\left(x-y\right)}^7}{x^2}=1$

Consider the curve defined by the equation

$\frac{1}{\sqrt[3]{x^3-y^3}}=\frac{2\mathrm{\pi }}{3}$ 

Use implicit differentiation to find $\frac{dx}{dy}$ in terms of $x$ and $y$.

By first rewriting the equation of the curve in the form $y=f\left(x\right)$:

(i) Determine the coordinates of the point on the curve where  $\frac{dy}{dx}=0$.

(ii) Explain why $f\left(x\right)$ is an increasing function on all intervals $\left[a,b\right]$ for which the interval $]a,b[$ does not include the $x$-coordinate of the point identified in part (b)(i).

(iii) Describe the asymptotic behaviour of the curve as $x\to \pm \infty$ .

After setting up a firework rocket on a stretch of level ground, the firework engineer lights the fuse and steps back to a safe distance of 10 metres from the rocket.  The rocket then begins to ascend vertically into the air at a constant velocity of 64 metres per second. 

Let $D$ be the distance, in metres, between the rocket and the point on the ground where the engineer is standing at time $t$ seconds after the rocket takes off.  Let $h$ be the height, in metres, of the rocket above the ground at time $t$ seconds.

Write an expression for $D$ in terms of $h$ only.

Use implicit differentiation to show that

 $\frac{dD}{dt}=\frac{64h}{\sqrt{h^2+100}}$

Find

(i) the rate at which the distance between the rocket and the point where the engineer is standing is increasing 1.56 seconds after the rocket takes off.

(ii) the height of the rocket above the ground at the moment when the distance between the rocket and the point where the engineer is standing is increasing at a rate of $10{\mathrm{ms}}^{-1}$ .

(i) Describe the mathematical behaviour of  $\frac{dD}{dt}$ as $h$ becomes large and interpret this in the context of the question.

(ii) Comment on the validity of the model for large values of $h$.

Quadrilateral $\mathrm{CDEF}$ represents a corral for unicorns.  There are fences along the four sides of the corral, as well as a straight fence across the middle connecting points $\mathrm{D}\mathrm{and}\mathrm{F}$.  Because of the way unicorns are trained, it is essential that triangles $\mathrm{CDF}$ and $\mathrm{DEF}$ be identical isosceles triangles, with $\mathrm{CD}=\mathrm{CF}=\mathrm{DE}=\mathrm{EF}$.  The length of side $\mathrm{DF}$, however, can vary.

Gonzolph is a unicorn trainer who is concerned about the high cost of unicorn fencing.  He would therefore like the total length of fencing, $P$, used in his corral to be the minimum possible for a given area, $A$, to be enclosed. 

Let  $\mathrm{DF}=2x\mathrm{m}$ and let  $\mathrm{CD}=y\mathrm{m}$.

By first finding the derivative $\frac{dA}{dx}$ in terms of $x$and $y$, show that for a given area the equation $\frac{dy}{dx}=\frac{2x^2-y^2}{xy}$ must be satisfied.

By considering the derivative $\frac{dP}{dx}$, show that when the length of fencing required to enclose a given area is the minimum possible then $x=\left(\frac{\sqrt{33}-1}{8}\right)y$.

Hence find the size of angle $\mathrm{F}\hat{\mathrm{C}}\mathrm{D}$ in a corral that minimises the amount of fencing required to enclose a given area.

Let  $f\left(x\right)={\left(ax+b\right)}^2$, where $a,b\in \mathrm{ℝ}$ are constants with $a\ne 0$ . 

By differentiating from first principles, show that $f\text{'}\left(x\right)=2a\left(ax+b\right)$.

Consider the function $f$ defined by

$f\left(x\right)={\mathrm{e}}^{-\frac{1}{2}x}\sin \left(\frac{\sqrt{3}}{2}x\right)$ 

By first calculating $f^{\text{'}}\left(x\right)$  and $f^{\text{'}\text{'}}\left(x\right)$,  show that $f^{\text{'}\text{'}\text{'}}\left(x\right)=f\left(x\right)$.

Write down the value of $f^{\left(29\right)}\left(0\right)$.

Find the derivative of the function
  $f\left(x\right)=\tan \left(\ln x\right)+\arctan \left({\mathrm{e}}^x\right)$.

Given that $g\left(x\right)={27}^{x+1}+\frac{1}{3}{\log }_3{x}^6$,  find $g\text{'}\left(x\right)$.  Simplify your answer as far as possible.

Let $h$ be the function defined by $h\left(x\right)=\arcsin \left(\cos x\right)$. Show that

$h\text{'}\left(x\right)=\{\begin{array}{ll}1,& \left(2k-1\right)\mathrm{\pi }<x<2k\mathrm{\pi }\\ -1,& 2k\mathrm{\pi }<x<\left(2\mathrm{k}+1\right)\mathrm{\pi }\\ \mathrm{undefined},& x=k\mathrm{\pi }\end{array}$ 

where $k\in \mathrm{\mathbb{Z}}$.

Use differentiation to show that $y=1+\tan x^3$ is a solution to the equation

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=18x^4\left(y^3-3y^2+4y-2\right)+6x\left(y^2-2y+2\right)$

Consider the curve defined by $y=\arccos \left(\ln x\right)$ ,  for values of $x$ satisfying $\frac{1}{\mathrm{e}}\le x\le \mathrm{e}$.

Show that

$y^n=\frac{1}{x^2\sqrt{1-{\ln }^{2}x}}\left(\frac{1-\ln x-{\ln }^{2}x}{1-{\ln }^{2}x}\right)$

Given that the curve has exactly one point of inflection, show that that point of inflection occurs when $x={\mathrm{e}}^{\frac{1}{\varphi }}$,  where $\varphi =\frac{1+\sqrt{5}}{2}$ is the so-called ‘golden ratio’.

Consider the function $f$ defined by $f\left(x\right)=8x^2+\cot 2x,0<x<\frac{\mathrm{\pi }}{2}$. 

The following diagram shows the graph of the curve $y=f\left(x\right)$:

The point marked A is the inflection point of the graph. 

Determine the exact coordinates of the point where the normal to the graph at point A  intersects the $y$-axis.

For each of the following, find $\frac{dy}{dx}$ by differentiating implicitly with respect to $x$.

$\frac{x^2}{3\sqrt{y}}+\frac{y^2}{4\sqrt{x}}=1$

$\sin \left(xy\right)={\left(x-2y\right)}^5$

$\sqrt{e^{-2x}-2e^{-x-y}+e^{-2y}}=\mathrm{\pi }$

A curve is described by the equation

 $\frac{x-2y}{{\left(xy\right)}^2}=k$ 

where $k\in \mathrm{ℝ}$ is a constant.

Use implicit differentiation to show that

$\frac{dy}{dx}=\frac{xy-4y^2}{2xy-2x^2}$

For a particular value of $k$, the curve goes through the point$\left(-1,-1\right)$. 

Find the value of $k$.

Find the equation of the

(i) tangent

(ii) normal

to the curve at the point $\left(2,-1\right)$.

Two observers, Pamela and Quinlan, are standing at points P and Q respectively watching a hot air balloon take off.  The balloon takes off from point O, which is in between points P and Q and is such that points P, O and Q all lie on a straight horizontal line.

Let  be the distance OP, and let $D_p$be the distance between point P and the balloon at any time $t$.  Similarly let $q$ be the distance OQ, and let $D_q$ be the distance between point Q and the balloon at any time $t$.  Let $h$ be the height of the balloon above the ground at any time $t$.  The balloon ascends vertically upwards, but its velocity during the ascent is not necessarily constant.  All distances are measured in metres, and all times in seconds.

Show that an expression for $\frac{d{D}_p}{dt}$ can be written solely in terms of $p$, $q$ and $D_q$.

Quinlan is standing a distance of 50 metres from where the balloon takes off.  At a certain moment in time, the balloon is at a distance of 112 metres from point Q and the distance between the balloon and point Q is increasing at a rate of 1.79 $\mathrm{m}{\mathrm{s}}^{-1}$.  At the same moment in time the distance between point P and the balloon is increasing at a rate of 1.05 $\mathrm{m}{\mathrm{s}}^{-1}$.

Use the above information and the results of part (a) to determine the distance that Pamela is standing from the point where the balloon takes off.

A third observer, Rhydderch, is standing at point R.  Point R is on the same side of point O as point P is, and it lies on the same horizontal line as points O, P and Q.  At the same moment described above, the distance between the balloon and point R is increasing at a rate of less than 0.8 metres per second. 

Find an inequality to express the minimum distance PR between the point where Rhydderch is standing and the point where Pamela is standing.

In the diagram below, $\mathrm{CDEFG}$ is a pentagon made up of a rectangle $\mathrm{CDFG}$, to one side of which an isosceles triangle $\mathrm{DEF}$ has been appended.  In addition sides $\mathrm{CD}$ and $\mathrm{FG}$ of the rectangle are the same length as the equal sides $\mathrm{DE}$ and $\mathrm{EF}$of the triangle.

The pentagon is intended to represent the cross-section of a new building, and the architect would like the area of the pentagon, A , to be the maximum possible for any given perimeter, $P$. 

Let $\mathrm{CG}=2x$units and let $\mathrm{DE}=y\mathrm{units}$.

By first finding the derivative $\frac{dP}{dx}$ in terms of $x$ and $y$, work out the value of the derivative $\frac{dy}{dx}$.

By considering the derivative $\frac{dA}{dx}$, show that when the area is maximal for a given perimeter the following equation must hold:

$20x^4-8x^{3}y-3x^2{y}^2+12x{y}^3-12y^4=0$

Hence determine

(i) the ratio of $x$ to $y$ (in the form $k:1$ for some $k$ to be determined) that gives the maximum area for a given perimeter, and

(ii) the maximum possible area for a pentagon of the above form with a perimeter of 100 metres.