IB Mathematics AHL 5.14 Implicit differentiation AA HL Paper 3

Question

If two functions $f(x)$ and $g(x)$ are differentiable, then their product is differentiable and the two functions satisfy the product rule: $(f(x)g(x))’ = f'(x)g(x) + g'(x)f'(x)$.

In this question, you will meet examples of pairs of differentiable functions, $f(x)$ and $g(x)$, that also satisfy $(f(x)g(x))’ = f'(x)g'(x)$.

In part (a), consider
$$f(x) = \frac{1}{(2-x)^2}$$
where $x \in \mathbb{R}$, $x \neq 2$, and $g(x) = x^2$ where $x \in \mathbb{R}$.

(a) (i) Find an expression for $f'(x)$.

(ii) Show that
$$f'(x)g'(x) = \frac{4x}{(2-x)^3}$$

(iii) Show that
$$f(x)g'(x) + g(x)f'(x) = \frac{4x}{(2-x)^3}$$

In parts (b) and (c), consider two non-constant functions, $f(x)$ and $g(x)$, where $f(x) > 0$ and $g(x) \neq g'(x)$.

(b) By rearranging the equation $f(x)g'(x) + g(x)f'(x) = f'(x)g'(x)$, show that
$$\frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x)-g(x)}$$

$$f(x) = Ae^{\int \frac{g'(x)}{g'(x)-g(x)}dx}$$

where A is an arbitrary positive constant.

The result from part (c) can be used to find pairs of functions, $f(x)$ and $g(x)$, which satisfy both of the following:

$$(f(x)g(x))’ = f(x)g'(x) + g(x)f'(x) \text{ and } (f(x)g(x))’ = f'(x)g'(x).$$

In parts (d) and (e), use the result in part (c) with A = 1.

(d) Consider $g(x) = xe^x$.

Find $f(x)$ such that $f(x)$ and $g(x)$ satisfy the above two equations.

(e) Consider $g(x) = \sin x + \cos x$.

Find $f(x)$ such that $f(x)$ and $g(x)$ satisfy the above two equations over the domain $0 < x < \pi$.

Give your answer in the form $f(x) = \sqrt{e^{h(x)}}$, where $h(x)$ is a function to be determined.

▶️Answer/Explanation

Detailed Solution

Part (a) (i) Given $f(x) = \frac{1}{(2 – x)^2}$ where $x \neq 2$,

OR \[ f(x) = (2 – x)^{-2} \]

Using the chain rule, let $u = 2 – x$, so $f(x) = u^{-2}$ and $\frac{du}{dx} = -1$. The derivative is:

\[ f'(x) = \frac{d}{dx} [u^{-2}] = -2 u^{-3} \cdot \frac{du}{dx} = -2 (2 – x)^{-3} \cdot (-1) = 2 (2 – x)^{-3} \]

Alternatively, express it as:

\[ f'(x) = \frac{2}{(2 – x)^3} \]……………..(eqn 1)

(ii) Given $g(x) = x^2$, finding its derivative:

\[ g'(x) = 2x \]……………..(eqn 2)

Now, taking the product $f'(x)g'(x)$ from equation 1 and 2:

\[ f'(x)g'(x) = \frac{2}{(2 – x)^3} \cdot 2x = \frac{4x}{(2 – x)^3} \]

\[ f'(x)g'(x) = \frac{4x}{(2 – x)^3} \]

(iii) To show $f(x)g'(x) + g(x)f'(x) = \frac{4x}{(2 – x)^3}$

Using the product rule, compute $(f(x)g(x))’$:

\[ (f(x)g(x))’ = f(x)g'(x) + g(x)f'(x) \]

Substitute the known functions and derivatives:

$f(x) = \frac{1}{(2 – x)^2}$
$g(x) = x^2$
$f'(x) = \frac{2}{(2 – x)^3}$
$g'(x) = 2x$

First term:

\[ f(x)g'(x) = \frac{1}{(2 – x)^2} \cdot 2x = \frac{2x}{(2 – x)^2} \]

Second term:

\[ g(x)f'(x) = x^2 \cdot \frac{2}{(2 – x)^3} = \frac{2x^2}{(2 – x)^3} \]

Add them:

\[ f(x)g'(x) + g(x)f'(x) = \frac{2x}{(2 – x)^2} + \frac{2x^2}{(2 – x)^3} \]

To combine, use a common denominator $(2 – x)^3$:

Rewrite the first term: $\frac{2x}{(2 – x)^2} = \frac{2x (2 – x)}{(2 – x)^3} = \frac{4x – 2x^2}{(2 – x)^3}$
Second term: $\frac{2x^2}{(2 – x)^3}$

\[ \frac{4x – 2x^2}{(2 – x)^3} + \frac{2x^2}{(2 – x)^3} = \frac{4x – 2x^2 + 2x^2}{(2 – x)^3} = \frac{4x}{(2 – x)^3} \] Hence proved

Part (b) To rearrange:

\[ f(x)g'(x) + g(x)f'(x) = f'(x)g'(x) \]

Move all terms to one side:

\[ f(x)g'(x) + g(x)f'(x) – f'(x)g'(x) = 0 \]

Factorize:

\[ f(x)g'(x) + f'(x)(g(x) – g'(x)) = 0 \]

\[ f'(x)(g(x) – g'(x)) = -f(x)g'(x) \]

Since $g(x) \neq g'(x)$, divide both sides by $f(x)(g'(x) – g(x))$ (noting $f(x) > 0$):

\[ \frac{f'(x)}{f(x)} = \frac{-f(x)g'(x)}{f(x)(g'(x) – g(x))} = \frac{-g'(x)}{g'(x) – g(x)} = \frac{g'(x)}{g(x) – g'(x)} \]

Thus:

\[ \frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x) – g(x)} \]

Part (c) Integrate both sides of:

\[ \frac{f'(x)}{f(x)} = \frac{g'(x)}{g'(x) – g(x)} \]

Left side:

\[ \int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C_1 \]

Since $f(x) > 0$, this is:

\[ \ln f(x) + C_1 \]

Right side:

This simplifies directly to:

\[ \int \frac{g'(x)}{g'(x) – g(x)} \, dx \]

Equate and solve:

\[ \ln f(x) + C_1 = \int \frac{g'(x)}{g'(x) – g(x)} \, dx + C_2 \]

\[ \ln f(x) = \int \frac{g'(x)}{g'(x) – g(x)} \, dx + C_2 – C_1 \]

\[ f(x) = e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx + C} = e^C \cdot e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx} \]

Since $f(x) > 0$, let $A = e^C > 0$:

\[ f(x) = A e^{\int \frac{g'(x)}{g'(x) – g(x)} \, dx} \] …………………….(eqn 3)

Part (d) Given $g(x) = x e^x$ and $A = 1$, find $f(x)$:

\[ g'(x) = e^x + x e^x = e^x (1 + x) \]

\[ g'(x) – g(x) = e^x (1 + x) – x e^x = e^x (1 + x – x) = e^x \]

\[ \frac{g'(x)}{g'(x) – g(x)} = \frac{e^x (1 + x)}{e^x} = 1 + x \]

\[ \int (1 + x) \, dx = x + \frac{x^2}{2} + C \]

From equation 3, \[ f(x) = 1 \cdot e^{x + \frac{x^2}{2} + C} = e^C e^{x + \frac{x^2}{2}} \]

Let $e^C = k > 0$, but since $A = 1$, adjust $C = 0$:

\[ f(x) = e^{x + \frac{x^2}{2}} \]

Part (e)

Given $g(x) = \sin x + \cos x$ over $0 < x < \pi$, and $f(x) = \sqrt{e^{h(x)}}$:

\[ g'(x) = \cos x – \sin x \]

\[ g'(x) – g(x) = (\cos x – \sin x) – (\sin x + \cos x) = 2 \cos x – 2 \sin x = 2 (\cos x – \sin x) \]

\[ \frac{g'(x)}{g'(x) – g(x)} = \frac{\cos x – \sin x}{2 (\cos x – \sin x)} = \frac{1}{2} \quad (\text{for } \cos x \neq \sin x) \]

At $x = \frac{\pi}{4}$, $\cos x = \sin x$, so adjust domain consideration. Generally:

\[ \int \frac{1}{2} \, dx = \frac{x}{2} + C \]

From equation 3 \[ f(x) = e^{\frac{x}{2} + C} = k e^{\frac{x}{2}}, \quad k = e^C \]

\[ f(x) = \sqrt{e^{\frac{x}{2} + C}} = e^{\frac{1}{2} (\frac{x}{2} + C)} \]

With $A = 1$, $C = 0$:

\[ f(x) = \sqrt{e^{\frac{x}{2}}} \]

So, $h(x) = \frac{x}{2}$.

————Markscheme—————–

Solution: –

(a) (i) attempts chain rule differentiation to find f(x)
$$f'(x)=\frac{2}{(2-x)^3}=(-1)(-2)(2-x)^{-3}$$

(ii)
$$g'(x)=2x$$
$$f'(x)g'(x)=(2(2-x)^{-3})(2x)\left [ =\frac{2(2x)}{(2-x)^3}\right ] (or equivalent)$$
$$=\frac{4x}{(2-x)^3} $$

(iii)

substitution f(x),g(x) and their g'(x),f'(x) into the given expression

$EITHER$
$$f(x)g'(x)+g(x)f'(x)=2x(2-x)^{-2}+2x^2(2-x)^{-3}$$

attempts to factorise their expression
$$=2x(2-x)^{-3}((2-x)+x)$$

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

attempts to form an expression with a common denominator

$$=\frac{2x(2-x)}{(2-x)^3}+\frac{2x^2}{(2-x)^3} = \left(\frac{4x-2x^2+2x^2}{(2-x)^3}\right)$$

THEN

$$=\frac{4x}{(2-x)^3}$$

(b) METHOD 1

$$f'(x)g'(x)-g(x)f'(x)=f(x)g'(x)$$
$$f'(x)g'(x)-g(x)f'(x)-f(x)g'(x)=0$$
$$f'(x)(g'(x)-g(x))=f(x)g'(x)$$
$$\frac{f'(x)}{f(x)}=\frac{g'(x)}{g'(x)-g(x)}$$

METHOD 2

$$g'(x)=\frac{f'(x)g'(x)}{f(x)}-\frac{g(x)f(x)}{f(x)}$$
$$g'(x)=\frac{f'(x)}{f(x)}(g(x)-g(x))$$
$$\frac{f'(x)}{f(x)}=\frac{g'(x)}{g'(x)-g(x)}$$

METHOD 3

$$g'(x)=\frac{f(x)g'(x)}{f'(x)}+g(x)$$
$$\frac{f(x)}{f'(x)}=\frac{g'(x)-g(x)}{g'(x)}$$
$$\frac{f'(x)}{f(x)}=\frac{g'(x)}{g'(x)-g(x)}$$

METHOD 4

$$\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}=1$$
$$\frac{f'(x)g(x)+g'(x)f(x)}{f(x)g(x)}=1$$
$$\frac{f'(x)}{f(x)}=\frac{g'(x)g(x)-g(x)}{g'(x)}$$
$$\frac{f'(x)}{f(x)}=\frac{g'(x)}{g'(x)-g(x)}$$

$$\ln f(x)+C=\int \frac{g'(x)}{g'(x)-g(x)} dx$$
$$f(x)=e\left ( ^{\int \frac{g'(x)}{g'(x)-g(x)} dx}e^-C \right )\left [f(x)e^{c}=e\left ( ^{\int \frac{g'(x)}{g'(x)-g(x)} dx} \right ),f(x)=e^{\left ( \int \frac{g'(x)}{g'(x)-g(x)}dx-c \right )} \right ]$$

THEN

$$f(x)=Ae^{\int \frac{g'(x)}{g'(x)-g(x)}dx}$$

METHOD 2

$$f'(x)-\frac{g'(x)}{g'(x)-g(x)}f(x)=0$$

Integrating factor:$$ e^{-\int \frac{g'(x)}{g'(x)-g(x)}dx}$$

$$\frac{d}{dx}\left [f(x)e^{-\int \frac{g'(x)}{g'(x)-g(x)}dx}\right ]=0$$

$$f(x)e^\left ( {-\int \frac{g'(x)}{g'(x)-g(x)}dx} \right )=A$$

$$f(x)=Ae^\left ({\int \frac{g'(x)}{g'(x)-g(x)}dx}\right )$$

(d) $g'(x) = xe^x + e^x \text{ (seen anywhere)}$

$\text{attempts to find an expression for } \frac{g'(x)}{g'(x) – g(x)}$

$=\frac{xe^x + e^x}{e^x}\left [ = \frac{e^x(x+1)}{e^x}\right ] =x+1(as e^{x}\neq 0)$

$\text{attempts to integrate their } \frac{g'(x)}{g'(x) – g(x)}$

$\int (x+1) dx = \frac{1}{2}x^2 + x \left ( + C \right )$

$f(x) = e^{\frac{1}{2}x^2 + x }$

(e) $g'(x) = \cos x – \sin x$ (seen anywhere)

$\text{attempts to find an expression for } \frac{g'(x)}{g'(x) – g(x)}$

$= \frac{\cos x – \sin x}{\cos x – \sin x- \sin x – \cos x} \left [= \frac{\sin x – \cos x}{2 \sin x}\right ]$

$= \frac{1}{2} – \frac{1}{2} \cot x \text{ (as } \sin x \neq 0 \text{)} \text{ OR }= \frac{1}{2}-\frac{1}{2} \frac{\cos x}{\sin x} \text{ (as } \sin x \neq 0 \text{)}$

$f(x) = e^{\int (\frac{1}{2} – \frac{1}{2} \cot x) dx}$

$\text{attempts to find the indefinite integral of } (\pm k) \cot x \text{ OR }(\pm k) \frac{\cos x}{\sin x}$

$\int \left( \frac{1}{2} – \frac{1}{2} \cot x \right) dx = \frac{x}{2} – \frac{1}{2} \ln |\sin x|( + C)\left [ =\frac{1}{2}\left ( x-1n\left|sin x \right|(+c) \right ) \right ]$

$f(x) = e^{\frac{x}{2}} e^{-\frac{1}{2} \ln |\sin x|}( e^C)$

$=e^{\frac{x}{2}}e^{1n\sqrt{\frac{1}{sin x}}}(e^{c})\left [ =e^{\frac{x}{2}}e^{\frac{1}{2}1n(\frac{1}{sin x})} (e^{c}),=\sqrt{e^{x-1n(sin x)}}(e^{c}) \right ]$

$=e^{\frac{x}{2}}\sqrt{\frac{1}{sin x}}$

$= \sqrt{e^x \csc x}\left [ =\sqrt{\frac{e^{x}}{sin x}} \right ]\text{ (where } h(x) = \frac{1}{\sin x} \text{)}$

Question 1

(a) Topic-SL 5.11 Areas between curves.

(b) Topic-SL 5.11 Areas between curves.

(d) Topic-SL 5.11 Areas between curves.

(e) Topic-AHL 5.14 Implicit differentiation.

In this question, you will investigate curved surface areas and use calculus to derive key formulae used in geometry.
Consider the straight line from the origin, $y = mx$, where $0 \leq x \leq h$ and $m, h$ are positive constants.

When this line is rotated through $360^\circ$ about the $x$-axis, a cone is formed with a curved surface area $A$ given by:
$
A = 2\pi \int_0^h y \sqrt{1 + m^2} \, dx.
$
(a) Given that $m = 2$ and $h = 3$, show that $A = 18 \sqrt{5} \pi$.
(b) Now consider the general case where a cone is formed by rotating the line $y = mx$ where $0 \leq x \leq h$ through $360^\circ$ about the $x$-axis.
(i) Deduce an expression for the radius of this cone $r$ in terms of $h$ and $m$.
(ii) Deduce an expression for the slant height $l$ in terms of $h$ and $m$.
(iii Hence, by using the above integral, show that $A = \pi r l$.
Consider the semi-circle, with radius $r$, defined by $y = \sqrt{r^2 – x^2}$ where $-r \leq x \leq r$.

(c) Find an expression for $\frac{dy}{dx}$.
A differentiable curve $y = f(x)$ is defined for $x_1 \leq x \leq x_2$, and $y \geq 0$. When any such curve is rotated through $360^\circ$ about the $x$-axis, the surface formed has an area $A$ given by:
$
A = 2\pi \int_{x_1}^{x_2} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx.
$
(d) A sphere is formed by rotating the semi-circle $y = \sqrt{r^2 – x^2}$ where $-r \leq x \leq r$ through $360^\circ$ about the $x$-axis. Show by integration that the surface area of this sphere is $4\pi r^2$.
(e) Let $f(x) = \sqrt{r^2 – x^2}$ where $-r \leq x \leq r$.
The graph of $y = f(x)$ is transformed to the graph of $y = f(kx)$, $k > 0$. This forms a different curve, called a semi-ellipse.
(i) Describe this geometric transformation.
(ii) Write down the $x$-intercepts of the graph $y = f(kx)$ in terms of $r$ and $k$.
(iii) For $y = f(kx)$, find an expression for $\frac{dy}{dx}$ in terms of $x$, $r$, and $k$.
(iv) The semi-ellipse $y = f(kx)$ is rotated $360^\circ$ about the $x$-axis to form a solid called an ellipsoid.
Find an expression in terms of $r$ and $k$ for the surface area, $A$, of the ellipsoid.
Give your answer in the form
$
2\pi \int_{x_1}^{x_2} \sqrt{p(x)} \, dx,
$
where $p(x)$ is a polynomial.
(v) Planet Earth can be modeled as an ellipsoid. In this model:
the ellipsoid has an axis of rotational symmetry running from the North Pole to the South Pole.
the distance from the North Pole to the South Pole is $12 \, 714 \, \text{km}$.
the diameter of the equator is $12 \, 756 \, \text{km}$.
By choosing suitable values for $r$ and $k$, find the surface area of Earth in $\text{km}^2$ correct to 4 significant figures. Give your answer in the form $a \times 10^q$ where $1 \leq a < 10$ and $q \in \mathbb{Z}^+$.

▶️Answer/Explanation

$\textbf{1(a)}$
$
A = 2\pi m \sqrt{1 + m^2} \left[ \frac{x^2}{2} \right]_0^h
= 2\pi m \sqrt{1 + m^2} \left( \frac{h^2}{2} \right)
$
$
= 2\pi (2) \sqrt{5} \left[ \frac{x^2}{2} \right]_0^3
= 2\pi (2) \sqrt{5} \left( \frac{3^2}{2} \right)
$
$
= 18 \sqrt{5} \pi
$

$\textbf{1(b)}$
(i) $r = mh$
(ii) $l = \sqrt{h^2 + r^2}$
$l = \sqrt{h^2 + h^2 m^2} \quad (= h\sqrt{1 + m^2})$
(iii) $A = 2\pi \int_0^h mx\sqrt{1 + m^2} \, dx$
$
= 2\pi m \sqrt{1 + m^2} \left[ \frac{1}{2} x^2 \right]_0^h
$
At least one of the above two lines needs to be shown.
$
= \pi h^2 m \sqrt{1 + m^2} \quad (= \pi h m \times \sqrt{h^2 + h^2 m^2})
$
$
= \pi r l
$

$\textbf{1(c)}$
Attempts implicit differentiation on $y^2 = r^2 – x^2$ (or equivalent):
$
\frac{dy}{dx} = -\frac{x}{y}
$

$\textbf{1(d)}$
$
A = 2\pi \int_{-r}^{r} \sqrt{r^2 – x^2} \sqrt{1 + \left(-x\left(r^2 – x^2\right)^{-\frac{1}{2}}\right)^2} \, dx
$
$
= 2\pi \int_{-r}^{r} \sqrt{r^2 – x^2} \sqrt{1 + \frac{x^2}{r^2 – x^2}} \, dx
$
Attempts to perform valid algebraic simplification to form a definite integral in terms of $r$ only:
$
= 2\pi \int_{-r}^{r} r \, dx
$
$
= 2\pi r \left[ x \right]_{-r}^{r} \quad (= 2\pi r \left(r – (-r)\right))
$
$
= 4\pi r^2
$

$\textbf{1(e)}$
(i) horizontal compression A1
factor k (invariant line y- axis)
(ii) $\pm \frac{r}{k}$
(iii) Attempts to use the chain rule:
$
\frac{dy}{dx} = \frac{1}{2} \left(r^2 – (kx)^2\right)^{-\frac{1}{2}} \times (-k^2 2x)
$
$
= -k^2 x \left(r^2 – (kx)^2\right)^{-\frac{1}{2}} \quad \left(= \frac{-k^2 x}{\sqrt{r^2 – k^2 x^2}} \right)
$
(iv) $
A = 2\pi \int_{-\frac{r}{k}}^{\frac{r}{k}} \sqrt{r^2 – k^2 x^2} \sqrt{1 + \frac{k^4 x^2}{r^2 – k^2 x^2}} \, dx
$
$
= 2\pi \int_{-\frac{r}{k}}^{\frac{r}{k}} \sqrt{r^2 – k^2 x^2 + k^4 x^2} \, dx
$
$
= 2\pi \int_{-\frac{r}{k}}^{\frac{r}{k}} \sqrt{r^2 + (k^4 – k^2)x^2} \, dx
$
(v)

$
r = 6378 \, \text{(km)}
$

$
k = 1.00330\ldots \left( = \frac{6378}{6357} = \frac{2126}{2119} \right)
$
Attempts to form a definite integral for surface area:
$
A = 2\pi \int_{-6357}^{6357} \sqrt{6378^2 – \left(\frac{6378}{6357}\right)^2 x^2 + \left(\frac{6378}{6357}\right)^4 x^2} \, dx
$
$
= 510064226.3\ldots
$
$
= 5.101 \times 10^8 \, \text{(km}^2\text{)}
$

IB Mathematics AHL 5.14 Implicit differentiation AA HL Paper 3 | Exam Style Questions

IB Mathematics AHL 5.14 Implicit differentiation AA HL Paper 3

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