Calculus: derivatives

(a) $y = cx$

Reason: The differential equation is $\dfrac{dy}{dx}=\dfrac{y}{x}$, which is separable:

$\frac{dy}{y}=\frac{dx}{x}\;\Rightarrow\;\ln y=\ln x+\ln c\;\Rightarrow\;y=cx$

Check: If $y=cx$, then $\dfrac{dy}{dx}=c$ and $\dfrac{y}{x}=\dfrac{cx}{x}=c$. Both sides are equal, so $y=cx$ satisfies the equation.

The function to be differentiated is missing from the question text, so no option can be marked. The standard derivatives tested in this type of question are:

None of the options

For $y = x^2 - 3x + 2$, $\dfrac{dy}{dx} = 2x - 3$. At $x = 2$ the tangent slope is $2(2) - 3 = 1$, so the slope of the normal is $-\dfrac{1}{1} = -1$. This value is not among (a)–(d), so the printed curve or options appear to be mistyped.

(a) $\ln|x| + C$

Since $\dfrac{d}{dx}(\ln|x|) = \dfrac{1}{x}$, the antiderivative of $\dfrac{1}{x}$ is $\ln|x| + C$. The absolute value is needed because $\ln x$ is defined only for $x>0$, while $\dfrac{1}{x}$ is defined for all $x \neq 0$.

(a) $\dfrac{dy}{dx} = \dfrac{x + y}{x - y}$

A differential equation $\frac{dy}{dx}=f(x,y)$ is homogeneous if $f(tx,ty)=f(x,y)$, i.e. $f$ is of degree $0$. Here

$f(tx,ty)=\frac{tx+ty}{tx-ty}=\frac{t(x+y)}{t(x-y)}=\frac{x+y}{x-y}=f(x,y),$

so it is homogeneous. In the other options, $f$ is not homogeneous of degree $0$: for (b), $x+y$ is homogeneous of degree $1$; for (c), $x^2+y^2$ is homogeneous of degree $2$; and for (d), $e^{x+y}$ is not homogeneous of degree $0$ since $e^{t(x+y)}\neq e^{x+y}$ in general.

(a) $e^{\int P\,dx}$

For the linear equation $\frac{dy}{dx} + Py = Q$, the integrating factor is $e^{\int P\,dx}$, which makes the left side exact as $\frac{d}{dx}\left(y\,e^{\int P\,dx}\right)$.

The derivative of $\sin^{-1} x$ is $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}, \quad -1 < x < 1.$

(b) L-Hospital's rule: If $f(x)$ and $g(x)$ are differentiable functions near $x = a$ such that $\lim\limits_{x \to a} \dfrac{f(x)}{g(x)}$ takes an indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, and $g'(x) \neq 0$ near $a$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)},$ provided the limit on the right exists.

(c) The curve $y = f(x)$ has a tangent parallel to the y-axis at a point where the slope of the tangent is infinite, i.e. where $\frac{dx}{dy} = 0 \quad \text{or} \quad \frac{dy}{dx} \to \infty.$

(d) The integral of $\dfrac{1}{\sqrt{a^2 - x^2}}$ is $\int \frac{1}{\sqrt{a^2 - x^2}}\, dx = \sin^{-1}\frac{x}{a} + c.$

(e) The standard form of a first order linear differential equation is $\frac{dy}{dx} + P(x)\,y = Q(x),$ where $P(x)$ and $Q(x)$ are functions of $x$ alone (or constants).

a. Let

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

Put $u=\dfrac{2x}{1-x^2}$. Then

$$\frac{dy}{dx}=\frac{1}{1+u^2}\frac{du}{dx}$$

Now,

$$\frac{du}{dx}=\frac{2(1-x^2)-2x(-2x)}{(1-x^2)^2} =\frac{2(1+x^2)}{(1-x^2)^2}$$

and

$$1+u^2=1+\frac{4x^2}{(1-x^2)^2} =\frac{(1+x^2)^2}{(1-x^2)^2}$$

Therefore,

$$\frac{dy}{dx} =\frac{2(1+x^2)}{(1-x^2)^2}\cdot \frac{(1-x^2)^2}{(1+x^2)^2} =\frac{2}{1+x^2}$$ $$\boxed{\frac{d}{dx}\left[\tan^{-1}\left(\frac{2x}{1-x^2}\right)\right]=\frac{2}{1+x^2}}$$

where $x\neq \pm 1$.

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b.

$$\int \frac{dx}{x^2-a^2}$$

Using partial fractions,

$$\frac{1}{x^2-a^2} =\frac{1}{(x-a)(x+a)} =\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right)$$

So,

$$\int \frac{dx}{x^2-a^2} =\frac{1}{2a}\left[\int \frac{dx}{x-a}-\int \frac{dx}{x+a}\right]$$ $$=\frac{1}{2a}\left[\ln|x-a|-\ln|x+a|\right]+C$$ $$\boxed{\int \frac{dx}{x^2-a^2} =\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C}$$

where $a\neq 0$.

a.
Let the volume and surface area of a sphere of radius $r$ be

$V=\frac{4}{3}\pi r^3,\qquad S=4\pi r^2.$

The rate of change of volume with respect to surface area is

$\frac{dV}{dS}=\frac{dV/dr}{dS/dr}.$

Now,

$\frac{dV}{dr}=4\pi r^2,\qquad \frac{dS}{dr}=8\pi r.$

So,

$\frac{dV}{dS}=\frac{4\pi r^2}{8\pi r}=\frac r2.$

Hence, the required rate of change is

$\boxed{\frac{dV}{dS}=\frac r2}.$

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b.
Let

$I=\int \frac{dx}{x^2+a^2}.$

Put

$x=a\tan\theta \quad\Rightarrow\quad dx=a\sec^2\theta\,d\theta.$

Then,

$I=\int \frac{a\sec^2\theta\,d\theta}{a^2\tan^2\theta+a^2}$

$=\int \frac{a\sec^2\theta\,d\theta}{a^2(\tan^2\theta+1)}$

$=\frac1a\int \frac{\sec^2\theta}{\sec^2\theta}\,d\theta$

$=\frac1a\int d\theta$

$=\frac1a\theta+C.$

Since $x=a\tan\theta$, we have $\theta=\tan^{-1}\left(\frac xa\right)$. Therefore,

$\boxed{\int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac xa\right)+C}.$

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c.
Given differential equation:

$\frac{dy}{dx}+y\cot x=\cos x.$

This is linear in $y$ with

$P=\cot x,\qquad Q=\cos x.$

Integrating factor:

$I.F.=e^{\int \cot x\,dx}=e^{\ln(\sin x)}=\sin x.$

Multiplying the equation by $\sin x$,

$\sin x\frac{dy}{dx}+y\cos x=\sin x\cos x.$

So,

$\frac{d}{dx}(y\sin x)=\sin x\cos x.$

Integrating both sides,

$y\sin x=\int \sin x\cos x\,dx+C.$

Put $t=\sin x$, then $dt=\cos x\,dx$. Hence,

$\int \sin x\cos x\,dx=\int t\,dt=\frac{t^2}{2}=\frac{\sin^2 x}{2}.$

Therefore,

$y\sin x=\frac{\sin^2 x}{2}+C.$

Dividing by $\sin x$,

$\boxed{y=\frac12\sin x+C\operatorname{cosec}x}.$

This is the required general solution.

The function to be differentiated is missing from the question text, so no option can be marked. The standard derivatives tested in this type of question are:

(b) $y = 2x - 1$

For the curve $y = x^2$ at the point $(1,1)$: $\dfrac{dy}{dx} = 2x$, so the slope at $x = 1$ is $m = 2$. The tangent is $y - 1 = 2(x - 1)$, i.e. $y = 2x - 1$.

The equation and options are missing from the question text, so no option can be marked. The usual method for this type of question is to substitute each option into the equation, or to solve it directly. For example, for $\dfrac{dy}{dx}=\dfrac{y}{x}$ separate the variables:

$\frac{dy}{y}=\frac{dx}{x}\;\Rightarrow\;\ln|y|=\ln|x|+\ln c\;\Rightarrow\;y=cx.$

L'Hospital's Rule (Theorem):

If $f(x)$ and $g(x)$ are differentiable functions on an open interval containing $a$ (except possibly at $a$ itself), and if

$\lim_{x \to a} \frac{f(x)}{g(x)}$

takes an indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, then

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)},$

provided the limit on the right exists (or is $\pm\infty$).

Application: It is used to evaluate limits of indeterminate forms, e.g.

$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1.$

The function is missing from the question text. Standard results:

$\frac{d}{dx}(\sin x)=\cos x,\quad \frac{d}{dx}(\cos x)=-\sin x,\quad \frac{d}{dx}(\tan x)=\sec^2 x,\quad \frac{d}{dx}(e^x)=e^x,\quad \frac{d}{dx}(\ln x)=\frac1x.$

$\dfrac{dy}{dx}=\dfrac{x^2+y^2}{xy}$

A homogeneous differential equation is one that can be written in the form $\dfrac{dy}{dx}=F\!\left(\dfrac{y}{x}\right)$, i.e. the right-hand side is a function of $\dfrac{y}{x}$ only. Here $\dfrac{x^2+y^2}{xy}=\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{1}{(y/x)}+\dfrac{y}{x}$, which is purely a function of $\dfrac{y}{x}$.

The integrand is missing from the question text. Standard integrals:

$\int 1\,dx = x + C,\qquad \int x^n\,dx=\frac{x^{n+1}}{n+1}+C\ (n\neq -1),\qquad \int \frac1x\,dx=\ln|x|+C,\qquad \int e^x\,dx=e^x+C.$

For a function $y=f(x)$:

• Actual change: $\Delta y = f(x+\Delta x) - f(x)$
• Approximate change: $dy = f'(x)\,\Delta x = \dfrac{dy}{dx}\,\Delta x$

The expressions are missing from the question text. The standard 3-mark methods are shown with examples.

Product rule: $\dfrac{d}{dx}(x^2\sin x)=2x\sin x+x^2\cos x$.

Quotient rule: $\dfrac{d}{dx}\left(\dfrac{x}{x+1}\right)=\dfrac{(x+1)-x}{(x+1)^2}=\dfrac{1}{(x+1)^2}$.

Chain rule: $\dfrac{d}{dx}\left(e^{x^2}\right)=2x\,e^{x^2}$.

The expression is missing from the question text. A typical 2-mark item of this type is a limit or an integral, for example

$\lim_{x\to 0}\frac{\sin 3x}{x}=3\lim_{x\to0}\frac{\sin 3x}{3x}=3\times 1=3.$

The rate of increase of the radius and the radius itself are missing from the question text. The method is as follows.

Volume of a sphere: $V=\dfrac43\pi r^3$. Differentiating with respect to time $t$,

$\frac{dV}{dt}=4\pi r^2\,\frac{dr}{dt}.$

Substitute the given $r$ and $\dfrac{dr}{dt}$. For example, if $\dfrac{dr}{dt}=2$ cm/s and $r=10$ cm, then $\dfrac{dV}{dt}=4\pi(100)(2)=800\pi\approx 2513$ cm$^3$/s.

The expression is missing from the question text. A standard 3-mark evaluation of this type is shown.

Evaluate: $\displaystyle\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$

Let $I=\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx \quad \cdots (1)$

Using $\displaystyle\int_0^{a}f(x)\,dx=\int_0^{a}f(a-x)\,dx$ with $a=\dfrac{\pi}{2}$,

$I=\int_0^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx \quad \cdots (2)$

Adding (1) and (2),

$2I=\int_0^{\pi/2}1\,dx=\frac{\pi}{2}\quad\Rightarrow\quad \boxed{I=\frac{\pi}{4}}$

The differential equation is missing from the question text. The method is to identify the type first, then solve.

• Variables separable: $\dfrac{dy}{dx}=f(x)g(y)$, so $\displaystyle\int\frac{dy}{g(y)}=\int f(x)\,dx$.
• Homogeneous: $\dfrac{dy}{dx}=F\!\left(\dfrac yx\right)$, so put $y=vx$.
• Linear: $\dfrac{dy}{dx}+Py=Q$, with integrating factor $e^{\int P\,dx}$.

Example (separable): $\dfrac{dy}{dx}=\dfrac{y}{x}$ gives $\dfrac{dy}{y}=\dfrac{dx}{x}$, so $\ln|y|=\ln|x|+\ln c$ and $y=cx$.

(a) $\dfrac{1}{\sqrt{1-x^2}}$

The derivative of $\sin^{-1}x$ is $\dfrac{1}{\sqrt{1-x^2}}$ (for $|x|<1$). The other options are the derivatives of $\cos^{-1}x$ (option b) and of $\tan^{-1}x$ (option c).

(c) 3

Since $\displaystyle \lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} 3\cdot \frac{\sin 3x}{3x} = 3 \times 1 = 3$.

(a) $x + 2y = 3$

For $y=x^2$, $\frac{dy}{dx}=2x$, so at $(1,1)$ the tangent slope is $2$. Hence the normal slope is $-\frac12$, giving $y-1=-\frac12(x-1)$, i.e. $x+2y=3$.

The order of a differential equation is the order of the highest derivative present.

Here the highest derivative is $\dfrac{d^2y}{dx^2}$, which is of order 2.

Order = 2

The derivative of $\sin^{-1}x$ with respect to $x$ is

$\frac{d}{dx}\left(\sin^{-1}x\right)=\frac{1}{\sqrt{1-x^2}}, \qquad -1<x<1.$

An exact differential equation is one of the form $M(x,y)\,dx + N(x,y)\,dy = 0$ where $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$.

Example: $\;2xy\,dx + (x^2 + 2y)\,dy = 0$

Here $M = 2xy$ and $N = x^2 + 2y$, so $\frac{\partial M}{\partial y} = 2x = \frac{\partial N}{\partial x}$ hence the equation is exact.

The integral of $\frac{1}{x}$ is:

$\int \frac{1}{x}\,dx = \ln|x| + C$

where $C$ is the constant of integration.

L'Hospital's rule: If $\lim_{x \to a} \dfrac{f(x)}{g(x)}$ takes the indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, then

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)},$

provided the limit on the right exists (here $f$ and $g$ are differentiable near $a$ with $g'(x) \neq 0$).

Standard result:

$\int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C$

Derivation (substitution): Put $x = a\tan\theta$, so $dx = a\sec^2\theta\,d\theta$ and $x^2 + a^2 = a^2\sec^2\theta$.

$\int \frac{1}{x^2+a^2}dx = \int \frac{a\sec^2\theta}{a^2\sec^2\theta}d\theta = \frac{1}{a}\int d\theta = \frac{1}{a}\theta + C = \frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right) + C$

Given: $\dfrac{dy}{dx} + y\cot x = \csc x$

This is linear in $y$ with $P = \cot x$, $Q = \csc x$.

Integrating factor: $\text{I.F.} = e^{\int \cot x\,dx} = e^{\ln(\sin x)} = \sin x$

Multiplying throughout by $\sin x$: $\sin x\frac{dy}{dx} + y\cos x = \sin x \cdot \csc x$ $\frac{d}{dx}(y\sin x) = 1$

Integrating both sides: $y\sin x = x + C$

$\boxed{y = (x + C)\csc x = \frac{x+C}{\sin x}}$

Given: cone with base radius $R=5$ cm and height $H=15$ cm; $\dfrac{dV}{dt}=5$ cm$^3$/s. Find $\dfrac{dh}{dt}$ when $h=10$ cm.

By similar triangles, the water surface radius $r$ at depth $h$ satisfies $\frac{r}{h}=\frac{5}{15}\;\Rightarrow\; r=\frac h3.$

Volume of water: $V=\frac13\pi r^2h=\frac13\pi\cdot\frac{h^2}{9}\cdot h=\frac{\pi h^3}{27}.$

Differentiating with respect to $t$: $\frac{dV}{dt}=\frac{\pi h^2}{9}\,\frac{dh}{dt}.$

Substituting $h=10$ and $\dfrac{dV}{dt}=5$: $5=\frac{100\pi}{9}\,\frac{dh}{dt}\;\Rightarrow\;\frac{dh}{dt}=\frac{9}{20\pi}\approx 0.143\text{ cm/s}.$

The water level is rising at about $0.143$ cm/s.

An anti-derivative is needed to undo differentiation, which is exactly what solving a differential equation requires: we are given a derivative and must recover the original function.

Justification: If $\dfrac{dy}{dx}=f(x)$, then solving means finding $y=\displaystyle\int f(x)\,dx$, i.e. an anti-derivative of $f(x)$.

Example: Solve $\dfrac{dy}{dx}=2x$.

$y=\int 2x\,dx=x^2+C$

Here $x^2+C$ is the anti-derivative of $2x$, and it gives the general solution of the differential equation. Without the anti-derivative, the original function $y$ cannot be recovered from its derivative.

A first-order, first-degree differential equation is homogeneous when the derivative can be expressed as a function of $\frac{y}{x}$ only, i.e.

$$\frac{dy}{dx}=f\left(\frac{y}{x}\right).$$

This happens when the coefficients of $dx$ and $dy$ are homogeneous functions of the same degree.

Example:

$$\frac{dy}{dx}=\frac{x^2+y^2}{xy}.$$

Here,

$$\frac{x^2+y^2}{xy} =\frac{x^2\left(1+\frac{y^2}{x^2}\right)}{x^2\left(\frac{y}{x}\right)} =\frac{1+\left(\frac{y}{x}\right)^2}{\frac{y}{x}},$$

so the equation is homogeneous.

Solution: Put

$$y=vx \quad\Rightarrow\quad \frac{dy}{dx}=v+x\frac{dv}{dx}.$$

Then the equation becomes

$$v+x\frac{dv}{dx}=\frac{1+v^2}{v} =v+\frac1v.$$

Hence,

$$x\frac{dv}{dx}=\frac1v$$ $$v\,dv=\frac{dx}{x}.$$

Integrating,

$$\int v\,dv=\int \frac{dx}{x}$$ $$\frac{v^2}{2}=\ln|x|+C.$$

Substituting $v=\frac{y}{x}$,

$$\frac{y^2}{2x^2}=\ln|x|+C.$$

Therefore,

$$\boxed{y^2=2x^2(\ln|x|+C)}$$

or

$$\boxed{y=\pm x\sqrt{2(\ln|x|+C)}}.$$

Thus the given condition $\frac{dy}{dx}=f\left(\frac{y}{x}\right)$ correctly characterizes a homogeneous first-order, first-degree differential equation.

The function to be differentiated and most options are missing from the question text. The derivative rules that apply to the visible option $2\cosh(x/4)$ are:

$\frac{d}{dx}\left[\sinh\frac x4\right]=\frac14\cosh\frac x4,\qquad \frac{d}{dx}\left[\tanh\frac x4\right]=\frac14\operatorname{sech}^2\frac x4.$

So $2\cosh(x/4)$ is the derivative of $8\sinh(x/4)$.

The integral and options are missing from the question text, so no option can be marked. Common standard integrals for this type of question:

$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac xa+C,\qquad \int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\frac xa+C,\qquad \int\frac{dx}{x^2-a^2}=\frac1{2a}\ln\left|\frac{x-a}{x+a}\right|+C.$

a. Theorem: Lagrange's Mean Value Theorem, since $f(x)$ is a curve on $[a,b]$ with a tangent at an interior point $C(x_1,y_1)$ parallel to the chord joining the end points. If $f(a)=f(b)$ the tangent at $C$ is horizontal and the theorem reduces to Rolle's theorem.

b. Conditions:

1. $f(x)$ is continuous on the closed interval $[a,b]$.
2. $f(x)$ is differentiable on the open interval $(a,b)$.

Then there is at least one $c\in(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}.$ (For Rolle's theorem, add the condition $f(a)=f(b)$, which gives $f'(c)=0$.)

c. $f'(c)$ represents the slope of the tangent to the curve at $x=c$, which equals the slope of the chord joining $(a,f(a))$ and $(b,f(b))$.

d. Tangent at $C(x_1,y_1)$: $y-y_1=f'(x_1)\,(x-x_1)$

e. Normal at $C(x_1,y_1)$: $y-y_1=-\frac{1}{f'(x_1)}\,(x-x_1),\qquad f'(x_1)\neq0,$ or equivalently $(x-x_1)+f'(x_1)(y-y_1)=0$.

The equation or expression is missing from the question text. For a 2-mark "Solve" item the usual form is a first-order equation, for example

$\frac{dy}{dx}=2x\;\Rightarrow\;y=\int 2x\,dx=x^2+C.$