Chapter 12 ยท Limits and Derivatives

12.3 The Derivative from First Principles

The derivative $f'(a)=\displaystyle\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$ is the instantaneous rate of change of $f$ at $a$ and the slope of the tangent to its graph. Compute it from first principles for powers, reciprocals, square roots, $\sin x$ and $\cos x$.

Draw a smooth curve and pick a point on it. Through that point there is exactly one straight line that "just touches" the curve there, running in the same direction as the curve โ€” the tangent line. For a circle you know how to draw it: perpendicular to the radius. For a parabola, a sine wave or the graph of a stone's height against time, there is no radius to help. How steep is the tangent? Where does it go?

In Section 12.1 we answered a question that turns out to be the same: the instantaneous speed of a falling stone. We took the average rate of change over a short interval, $\tfrac{f(a+h)-f(a)}{h}$ โ€” the slope of a chord, or secant line, through two points of the graph โ€” and let $h\to0$. In this section we give that limit its name, the derivative, and learn to compute it straight from the definition ("from first principles"). The results for $x^2$, $x^3$, $\tfrac1x$, $\sqrt x$, $\sin x$ and $\cos x$ found here are the raw material for the quick rules of Section 12.4.

The Derivative at a Point

๐Ÿ“˜ Definition 12.3 โ€” Derivative at a point

Let $f$ be a real function defined on an open interval containing the point $a$. The derivative of $f$ at $a$ is $$f'(a)=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h},$$ provided this limit exists (as a real number). When it exists we say that $f$ is differentiable at $a$. Putting $x=a+h$ gives the equivalent form $f'(a)=\displaystyle\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}$.

The fraction $\tfrac{f(a+h)-f(a)}{h}$ is called the difference quotient. It always has the form $\tfrac00$ at $h=0$, so finding a derivative always means simplifying before letting $h\to0$. The derivative has two meanings, and it pays to keep both in mind.

๐Ÿ“ As a rate of change

The difference quotient is the average rate of change of $f$ over $[a,a+h]$. Its limit $f'(a)$ is the instantaneous rate of change of $f$ at $a$. If $s(t)$ is position, $s'(t)$ is the velocity; if $V(t)$ is the volume in a tank, $V'(t)$ is the flow rate.

๐Ÿ“ As a slope

The difference quotient is the slope of the secant line through $P(a,f(a))$ and $Q(a+h,f(a+h))$. As $h\to0$, $Q$ slides along the curve to $P$ and the secant turns into the tangent at $P$. So $f'(a)$ is the slope of the tangent, and the tangent line is $$y-f(a)=f'(a)(x-a).$$

The tangent-line equation is the point-slope form of a line from Section 9.2, with the point $(a,f(a))$ and slope $m=f'(a)$.

โœ๏ธ Example 1 โ€” The derivative of $x^2$ at a point, and the tangent
Let $f(x)=x^2$. (a) Find $f'(3)$ from the definition. (b) Find the equation of the tangent to $y=x^2$ at $(3,9)$. (c) Where does this tangent cross the $x$-axis?
(a) $f(3+h)-f(3)=(9+6h+h^2)-9=6h+h^2$. For $h\ne0$, $\dfrac{6h+h^2}{h}=6+h$, so $f'(3)=\displaystyle\lim_{h\to0}(6+h)=6$.
(b) Point $(3,9)$, slope $6$: $y-9=6(x-3)$, i.e. $y=6x-9$.
(c) $0=6x-9$ gives $x=1.5$: the tangent meets the $x$-axis at $(1.5,0)$.
Check The line touches the parabola only at $x=3$: $x^2=6x-9$ means $(x-3)^2=0$, a double root โœ“.
โœ๏ธ Example 2 โ€” Velocity of a ball thrown upwards
A ball is thrown straight up. Its height after $t$ seconds is $y(t)=20t-5t^2$ metres. Find its velocity at $t=1\ \text{s}$ and at $t=3\ \text{s}$, and explain the signs.
t = 1 $y(1)=15$ and $y(1+h)=20+20h-5(1+2h+h^2)$ $=15+10h-5h^2$. So $\dfrac{y(1+h)-y(1)}{h}=10-5h\to10$. The velocity is $10\ \text{m/s}$.
t = 3 $y(3)=15$ and $y(3+h)=60+20h-5(9+6h+h^2)$ $=15-10h-5h^2$. The quotient is $-10-5h\to-10$: velocity $-10\ \text{m/s}$.
Meaning At both times the ball is $15\ \text{m}$ up. At $t=1$ the height is increasing (moving up at $10\ \text{m/s}$); at $t=3$ it is decreasing (falling at $10\ \text{m/s}$). The sign of the derivative tells the direction.
โœ๏ธ Example 3 โ€” A point with no derivative
Show that $f(x)=|x|$ is not differentiable at $x=0$, and find $f'(2)$ and $f'(-2)$.
At 0 $\dfrac{f(0+h)-f(0)}{h}=\dfrac{|h|}{h}$, which is $1$ for $h\gt0$ and $-1$ for $h\lt0$. The one-sided limits are $1$ and $-1$, so the limit โ€” and hence $f'(0)$ โ€” does not exist.
Picture Secants from the right all have slope $1$, secants from the left all have slope $-1$. At the corner there is no single tangent direction.
At ยฑ2 Near $x=2$, $f(x)=x$, so the quotient is $\tfrac{(2+h)-2}{h}=1$: $f'(2)=1$. Near $x=-2$, $f(x)=-x$, so $f'(-2)=-1$.
Interactive 1 โ€” From Secant to Tangent

$P$ is fixed on the curve and $Q$ is the point at $x=a+h$. The red line is the secant $PQ$, with its run $h$ and rise $f(a+h)-f(a)$ shown; the blue dashed line is the tangent at $P$. Press Play to let $h$ shrink towards $0$ (alternately from the right and from the left), or set it yourself. Watch the secant slope converge on $f'(a)$.

1
$h$โ€”
secant slopeโ€”
$f'(a)$โ€”
differenceโ€”

The Derivative as a Function

In Example 1 we found $f'(3)=6$ for $f(x)=x^2$. Repeating the work at $a=1$, $a=-2$, $a=10$โ€ฆ would be wasteful. Instead we do it once, at a general point $x$.

๐Ÿ“˜ Definition 12.4 โ€” The derivative function (first principles)

The derivative of $f$ is the function $f'$ whose value at $x$ is $$f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h},$$ defined at every $x$ where this limit exists. Finding $f'(x)$ directly from this limit is called differentiating from first principles. If $y=f(x)$, the derivative is also written $$f'(x)=\dfrac{dy}{dx}=\dfrac{d}{dx}f(x)=y'.$$

The notation $\tfrac{dy}{dx}$ is due to Leibniz: it reminds us that the derivative is the limit of $\tfrac{\Delta y}{\Delta x}$, a change in $y$ divided by a change in $x$. It is a single symbol, not a fraction you may cancel. The value at a point is written $f'(a)$ or $\left.\tfrac{dy}{dx}\right|_{x=a}$.

๐Ÿงญ Problem-solving strategy โ€” Differentiating from first principles
  1. Write $f(x+h)$ by replacing every $x$ in the formula by $(x+h)$ โ€” with brackets.
  2. Form the difference $f(x+h)-f(x)$ and simplify it until every term contains the factor $h$. Tools: expanding (binomial theorem), a common denominator for fractions, rationalising for square roots, a sum or difference formula for trigonometric functions.
  3. Divide by $h$ and cancel it (allowed because $h\ne0$).
  4. Only now let $h\to0$, using substitution or a standard limit such as $\tfrac{\sin h}{h}\to1$.
โœ๏ธ Example 4 โ€” Constants, lines and powers
Find from first principles the derivatives of (a) $f(x)=c$, a constant (b) $f(x)=mx+b$ (c) $f(x)=x^2$ (d) $f(x)=x^3$.
(a) $\dfrac{c-c}{h}=0$ for every $h\ne0$, so $f'(x)=0$. A horizontal line has slope $0$ everywhere.
(b) $\dfrac{m(x+h)+b-(mx+b)}{h}=\dfrac{mh}{h}=m$, so $f'(x)=m$. A line is its own tangent; its slope is $m$ everywhere.
(c) $(x+h)^2-x^2=2xh+h^2$, so the quotient is $2x+h\to2x$: $\dfrac{d}{dx}(x^2)=2x$. (At $x=3$ this gives $6$, as in Example 1.)
(d) $(x+h)^3-x^3=3x^2h+3xh^2+h^3$, so the quotient is $3x^2+3xh+h^2\to3x^2$: $\dfrac{d}{dx}(x^3)=3x^2$.
โœ๏ธ Example 5 โ€” Reciprocals and square roots
Find from first principles the derivatives of (a) $f(x)=\dfrac1x$, $x\ne0$ (b) $f(x)=\sqrt x$, $x\gt0$ (c) $f(x)=\dfrac{1}{x^2}$, $x\ne0$.
(a) $\dfrac1{x+h}-\dfrac1x=\dfrac{x-(x+h)}{x(x+h)}=\dfrac{-h}{x(x+h)}$. Dividing by $h$: $\dfrac{-1}{x(x+h)}\to-\dfrac1{x^2}$.
(b) Rationalise: $\sqrt{x+h}-\sqrt x=\dfrac{(x+h)-x}{\sqrt{x+h}+\sqrt x}$ $=\dfrac{h}{\sqrt{x+h}+\sqrt x}$. Dividing by $h$: $\dfrac1{\sqrt{x+h}+\sqrt x}\to\dfrac1{2\sqrt x}$.
(c) $\dfrac1{(x+h)^2}-\dfrac1{x^2}=\dfrac{x^2-(x+h)^2}{x^2(x+h)^2}$ $=\dfrac{-h(2x+h)}{x^2(x+h)^2}$. Dividing by $h$: $\dfrac{-(2x+h)}{x^2(x+h)^2}\to\dfrac{-2x}{x^4}=-\dfrac{2}{x^3}$.
Pattern $x^2\to2x$, $x^3\to3x^2$, $x^{-1}\to-x^{-2}$, $x^{-2}\to-2x^{-3}$, $x^{1/2}\to\tfrac12x^{-1/2}$: in every case "bring the power down and reduce it by $1$". Section 12.4 proves this for every integer power.
Interactive 2 โ€” The First-Principles Stepper

Choose a function and step through the four moves of the strategy (or press Play). The graph shows the difference quotient $q(h)=\tfrac{f(x_0+h)-f(x_0)}{h}$ as a function of $h$ at the point $x_0$ you choose: it is undefined at $h=0$ โ€” a hole โ€” and the height of the hole is $f'(x_0)$.

1
stepโ€”
$q(-0.01)$โ€”
$q(0.01)$โ€”
$f'(x_0)$โ€”

Derivatives of $\sin x$ and $\cos x$

For trigonometric functions, step 2 of the strategy uses the sum formulae of Section 3.3, and step 4 uses the two standard limits of Section 12.2. That is why we needed $\tfrac{\sin h}{h}\to1$ and $\tfrac{1-\cos h}{h}\to0$.

๐Ÿ“ Theorem 12.8 โ€” Derivatives of sine and cosine

For every real $x$ (in radians), $$\dfrac{d}{dx}(\sin x)=\cos x,\qquad\dfrac{d}{dx}(\cos x)=-\sin x.$$

Proof. By the sum formula, $\sin(x+h)=\sin x\cos h+\cos x\sin h$, so $$\begin{aligned}\dfrac{\sin(x+h)-\sin x}{h}&=\cos x\cdot\dfrac{\sin h}{h}\\&\quad-\sin x\cdot\dfrac{1-\cos h}{h}.\end{aligned}$$ Here $x$ is fixed and only $h$ varies. As $h\to0$ the first fraction tends to $1$ and the second to $0$ (Theorem 12.7), so the limit is $\cos x\cdot1-\sin x\cdot0=\cos x$.

Similarly $\cos(x+h)=\cos x\cos h-\sin x\sin h$, so $$\begin{aligned}\dfrac{\cos(x+h)-\cos x}{h}&=-\cos x\cdot\dfrac{1-\cos h}{h}\\&\quad-\sin x\cdot\dfrac{\sin h}{h},\end{aligned}$$ which tends to $-\cos x\cdot0-\sin x\cdot1=-\sin x$. $\blacksquare$

The graphs agree. $\sin x$ is steepest upwards at $x=0$, where $\cos0=1$; it is flat at its peak $x=\tfrac\pi2$, where $\cos\tfrac\pi2=0$; and it falls fastest at $x=\pi$, where $\cos\pi=-1$. The radian measure matters: in degrees, the slope of $\sin x^\circ$ at $0$ would be $\tfrac{\pi}{180}$, not $1$.

โœ๏ธ Example 6 โ€” Trigonometric functions from first principles
Find from first principles the derivatives of (a) $\sin2x$ (b) $\tan x$, for $\cos x\ne0$.
(a) By the formula $\sin A-\sin B=2\cos\tfrac{A+B}2\sin\tfrac{A-B}2$ (Section 3.4): $\sin(2x+2h)-\sin2x=2\cos(2x+h)\sin h$.
(a) Limit $\dfrac{2\cos(2x+h)\sin h}{h}=2\cos(2x+h)\cdot\dfrac{\sin h}{h}$ $\to2\cos2x\cdot1=2\cos2x$.
(b) $\tan(x+h)-\tan x$ $=\dfrac{\sin(x+h)\cos x-\cos(x+h)\sin x}{\cos(x+h)\cos x}$ $=\dfrac{\sin h}{\cos(x+h)\cos x}$, using $\sin(A-B)$ with $A=x+h$, $B=x$.
(b) Limit $\dfrac{\sin h}{h}\cdot\dfrac{1}{\cos(x+h)\cos x}\to1\cdot\dfrac{1}{\cos^2x}=\sec^2x$. So $\dfrac{d}{dx}(\tan x)=\sec^2x$.
Check At $x=\tfrac\pi4$: $\sec^2\tfrac\pi4=2$, and $\tfrac{\tan(0.786398)-1}{0.001}\approx2.002$ โœ“.
โš ๏ธ Common mistakes with first principles

1. Writing $f(x+h)=f(x)+h$. For $f(x)=x^2$, $f(x+h)=(x+h)^2=x^2+2xh+h^2$, not $x^2+h$. Replace every $x$ by $(x+h)$.

2. Putting $h=0$ too early. Setting $h=0$ in $\tfrac{(x+h)^2-x^2}{h}$ gives $\tfrac00$. You must cancel $h$ first; only then let $h\to0$.

3. "Cancelling" $h$ that is not a factor. In $\tfrac{2xh+h^2}{h}$, $h$ is a factor of both terms, giving $2x+h$. In $\tfrac{(x+h)^2}{h}$ it is not โ€” expand and subtract $f(x)$ first.

4. Using $f'(x)$ for the tangent instead of $f'(a)$. The tangent at $x=a$ has the number $f'(a)$ as its slope: $y-f(a)=f'(a)(x-a)$. Writing $y-9=2x(x-3)$ for the tangent to $y=x^2$ at $(3,9)$ gives a parabola, not a line.

5. Degrees. $\tfrac{d}{dx}\sin x=\cos x$ only for radians, because it rests on $\tfrac{\sin h}{h}\to1$.

6. Assuming every function has a derivative. At a corner (like $|x|$ at $0$) or a jump (like $[x]$ at an integer) the difference quotient has no limit, so there is no derivative.

Exercises

I
The derivative at a point (use the definition).
  • 1 $f(x)=x^2+1$; find $f'(2)$.
  • 2 $f(x)=5x-3$; find $f'(-1)$.
  • 3 $f(x)=2x^2-3x$; find $f'(1)$.
  • 4 $f(x)=x^3$; find $f'(-2)$.
  • 5 $f(x)=\dfrac1x$; find $f'(2)$.
  • 6 $f(x)=\sqrt x$; find $f'(9)$.
  • 7 Find the equation of the tangent to $y=x^2-4x$ at the point where $x=3$.
  • 8 Show that $f(x)=|x-1|$ is not differentiable at $x=1$.
1 $\dfrac{(2+h)^2+1-5}{h}=\dfrac{4h+h^2}{h}=4+h\to4$.
2 $\dfrac{5(-1+h)-3-(-8)}{h}=\dfrac{5h}{h}=5$.
3 $f(1)=-1$; $f(1+h)=2+4h+2h^2-3-3h$ $=-1+h+2h^2$. Quotient $1+2h\to1$.
4 $(-2+h)^3=-8+12h-6h^2+h^3$, so the quotient is $12-6h+h^2\to12$.
5 $\dfrac{\frac1{2+h}-\frac12}{h}=\dfrac{-1}{2(2+h)}\to-\dfrac14$.
6 $\dfrac{\sqrt{9+h}-3}{h}=\dfrac{1}{\sqrt{9+h}+3}\to\dfrac16$.
7 $f(3)=-3$; $\dfrac{(3+h)^2-4(3+h)+3}{h}=\dfrac{2h+h^2}{h}\to2$. Tangent: $y+3=2(x-3)$, i.e. $y=2x-9$.
8 $\dfrac{|1+h-1|-0}{h}=\dfrac{|h|}{h}$: right-hand limit $1$, left-hand limit $-1$. They differ, so $f'(1)$ does not exist (the graph has a corner at $(1,0)$).
II
Derivatives from first principles.
  • 1 $f(x)=7$
  • 2 $f(x)=3x+5$
  • 3 $f(x)=x^2-2x$
  • 4 $f(x)=x^3+x$
  • 5 $f(x)=\dfrac{1}{x+2}$, $x\ne-2$
  • 6 $f(x)=\dfrac2x$, $x\ne0$
  • 7 $f(x)=\sqrt{x+1}$, $x\gt-1$
  • 8 $f(x)=\dfrac{1}{2x+1}$, $x\ne-\tfrac12$
  • 9 $f(x)=x^4$ (use the binomial theorem)
Check: the difference quotient at small $h$ against your answer
1
$q(-0.001)$โ€”
$q(0.001)$โ€”
answer formula at $x$โ€”
1 $\tfrac{7-7}{h}=0$: $f'(x)=0$.
2 $\tfrac{3h}{h}=3$: $f'(x)=3$.
3 $(x+h)^2-2(x+h)-x^2+2x$ $=2xh+h^2-2h$; quotient $2x+h-2\to2x-2$.
4 Difference $3x^2h+3xh^2+h^3+h$; quotient $3x^2+3xh+h^2+1\to3x^2+1$.
5 $\dfrac1{x+h+2}-\dfrac1{x+2}=\dfrac{-h}{(x+h+2)(x+2)}$; quotient $\to-\dfrac{1}{(x+2)^2}$.
6 $\dfrac2{x+h}-\dfrac2x=\dfrac{-2h}{x(x+h)}$; quotient $\to-\dfrac2{x^2}$.
7 Rationalise: $\dfrac{\sqrt{x+h+1}-\sqrt{x+1}}{h}$ $=\dfrac{1}{\sqrt{x+h+1}+\sqrt{x+1}}$ $\to\dfrac{1}{2\sqrt{x+1}}$.
8 $\dfrac1{2x+2h+1}-\dfrac1{2x+1}$ $=\dfrac{-2h}{(2x+2h+1)(2x+1)}$; quotient $\to-\dfrac{2}{(2x+1)^2}$.
9 $(x+h)^4=x^4+4x^3h+6x^2h^2+4xh^3+h^4$ (Section 7.1). Quotient $4x^3+6x^2h+4xh^2+h^3\to4x^3$.
III
Trigonometric functions from first principles ($x$ in radians).
  • 1 $\cos3x$
  • 2 $\sin(x+1)$
  • 3 $x+\sin x$
  • 4 $\sin x\cos x$ (first write it as a single sine)
  • 5 $\sec x$, where $\cos x\ne0$
  • 6 $\cos^2x$ (first write it using $\cos2x$)
1 $\cos(3x+3h)-\cos3x$ $=-2\sin\left(3x+\tfrac{3h}2\right)\sin\tfrac{3h}2$. Quotient $=-3\sin\left(3x+\tfrac{3h}2\right)\cdot\dfrac{\sin(3h/2)}{3h/2}\to-3\sin3x$.
2 $\sin(x+1+h)-\sin(x+1)$ $=2\cos\left(x+1+\tfrac h2\right)\sin\tfrac h2$; quotient $\to\cos(x+1)$.
3 The quotient is $\dfrac hh+\dfrac{\sin(x+h)-\sin x}{h}\to1+\cos x$.
4 $\sin x\cos x=\tfrac12\sin2x$. By Example 6(a), the derivative is $\tfrac12\cdot2\cos2x=\cos2x$.
5 $\dfrac1{\cos(x+h)}-\dfrac1{\cos x}$ $=\dfrac{\cos x-\cos(x+h)}{\cos x\cos(x+h)}$ $=\dfrac{2\sin\left(x+\frac h2\right)\sin\frac h2}{\cos x\cos(x+h)}$. Dividing by $h$ and letting $h\to0$: $\dfrac{\sin x}{\cos^2x}=\sec x\tan x$.
6 $\cos^2x=\tfrac12+\tfrac12\cos2x$. The constant contributes $0$, and as in part 1, $\tfrac{d}{dx}\cos2x=-2\sin2x$. So the derivative is $-\sin2x$ $(=-2\sin x\cos x)$.
IV
Rates of change and tangents.
  • 1 A stone is thrown upwards and its height is $y=30t-5t^2$ metres after $t$ seconds. Find its velocity at a general time $t=a$ from first principles, then at $t=1$ and $t=4$. When is the velocity $0$, and how high is the stone then?
  • 2 Find the rate of change of the area $A=\pi r^2$ of a circle with respect to its radius when $r=5\ \text{cm}$. What do you notice about the answer?
  • 3 Find the rate of change of the volume $V=x^3$ of a cube with respect to its edge $x$ when $x=2\ \text{cm}$.
  • 4 A workshop's cost of making $x$ chairs is $C(x)=500+20x+0.05x^2$ rupees. Find $C'(x)$ from first principles and the marginal cost $C'(100)$. Compare it with the actual cost of the $101$st chair, $C(101)-C(100)$.
  • 5 Find the tangent lines to $y=\sin x$ at $x=0$ and to $y=\cos x$ at $x=\tfrac\pi2$.
1 $\dfrac{y(a+h)-y(a)}{h}=\dfrac{30h-5(2ah+h^2)}{h}$ $=30-10a-5h\to30-10a$. So $v(1)=20\ \text{m/s}$ (rising), $v(4)=-10\ \text{m/s}$ (falling). $v=0$ when $a=3\ \text{s}$; then $y=90-45=45\ \text{m}$, the highest point.
2 $\dfrac{\pi(r+h)^2-\pi r^2}{h}=\pi(2r+h)\to2\pi r$. At $r=5$: $10\pi\approx31.4\ \text{cm}^2$ per cm. The rate is the circumference: a thin ring of width $h$ adds about $2\pi r\cdot h$ to the area.
3 $V'(x)=3x^2$ (Example 4(d)); at $x=2$: $12\ \text{cm}^3$ per cm.
4 $\dfrac{C(x+h)-C(x)}{h}=20+0.05(2x+h)$ $\to20+0.1x$. $C'(100)=30$, i.e. โ‚น30 per chair. $C(101)-C(100)$ $=20+0.05(10201-10000)$ $=30.05$, i.e. โ‚น30.05 โ€” very close to the marginal cost.
5 $\tfrac{d}{dx}\sin x=\cos x$, so the slope at $0$ is $1$ and the tangent is $y=x$. $\tfrac{d}{dx}\cos x=-\sin x$, so the slope at $\tfrac\pi2$ is $-1$; the point is $\left(\tfrac\pi2,0\right)$ and the tangent is $y=\tfrac\pi2-x$.
๐ŸŽฏ Exam corner Board ยท 1 markBoard ยท 2 marksBoard ยท 3 marksBoard ยท Case studyJEE MainJEE Advanced

How this is examined: "Find the derivative of โ€ฆ from first principles" is a near-certain Board question (2โ€“3 marks) โ€” polynomials, $\tfrac1x$-type fractions, square roots and simple trigonometric functions, with every step of the limit shown. Rate-of-change case studies use the derivative as a velocity. JEE uses the derivative as a slope (tangent conditions) and tests differentiability at corners.

E1. Board ยท 1 mark For $f(x)=x^2$, the value of $\displaystyle\lim_{h\to0}\dfrac{f(-3+h)-f(-3)}{h}$ is (A) $-6$   (B) $6$   (C) $9$   (D) $-9$

Idea The limit is $f'(-3)$. $\dfrac{(9-6h+h^2)-9}{h}=-6+h\to-6$.
Trap (C) is the value $f(-3)$, not the slope; (B) drops the sign of $x=-3$.
Answer (A).

E2. Board ยท 2 marks Find the derivative of $f(x)=\dfrac{x}{x+1}$ from first principles.

Difference $\dfrac{x+h}{x+h+1}-\dfrac{x}{x+1}$ $=\dfrac{(x+h)(x+1)-x(x+h+1)}{(x+h+1)(x+1)}$ $=\dfrac{h}{(x+h+1)(x+1)}$. (1 mark)
Answer Dividing by $h$: $\dfrac{1}{(x+h+1)(x+1)}\to\dfrac{1}{(x+1)^2}$, for $x\ne-1$. (1 mark)

E3. Board ยท 3 marks Find the derivative of $x\sin x$ from first principles.

Split $(x+h)\sin(x+h)-x\sin x$ $=x[\sin(x+h)-\sin x]+h\sin(x+h)$. (1 mark)
Divide $\dfrac{(x+h)\sin(x+h)-x\sin x}{h}$ $=x\cdot\dfrac{2\cos\left(x+\frac h2\right)\sin\frac h2}{h}+\sin(x+h)$. (1 mark)
Answer As $h\to0$: $x\cos x\cdot1+\sin x$. So $\dfrac{d}{dx}(x\sin x)=\sin x+x\cos x$. (1 mark)

E4. Board ยท Case study On a test track a car is $s(t)=2t^2+3t$ metres past the start line $t$ seconds after it begins. (i) Find its average velocity over $[1,3]$. (1 mark) (ii) Find its velocity at $t=2$ from first principles. (2 marks) (iii) At what time is its velocity $15\ \text{m/s}$? (1 mark)

(i) $s(3)=27$, $s(1)=5$: $\dfrac{27-5}{2}=11\ \text{m/s}$. (1 mark)
(ii) $s(2)=14$; $s(2+h)=2(4+4h+h^2)+6+3h$ $=14+11h+2h^2$. Quotient $11+2h\to11\ \text{m/s}$. (For a quadratic, the average over an interval equals the velocity at its mid-point.) (2 marks)
(iii) In general $\dfrac{s(t+h)-s(t)}{h}=4t+3+2h\to4t+3$; $4t+3=15$ gives $t=3\ \text{s}$. (1 mark)

E5. JEE Main The tangent to $y=x^2-5x+6$ at the point $P$ is parallel to the line $y=3x+1$. Then $P$ is (A) $(4,2)$   (B) $(1,2)$   (C) $(4,3)$   (D) $(-1,12)$

Method $\tfrac{dy}{dx}=2x-5$ (from first principles, as in Exercise II.3). Parallel to slope $3$: $2x-5=3$, $x=4$, $y=16-20+6=2$.
Trap (B) lies on the curve but its slope is $2-5=-3$ โ€” the sign is wrong; (D) is on the curve with slope $-7$; (C) is not on the curve at all.
Answer (A).

E6. JEE Advanced Numerical answer. Let $f(x)=x|x|$. Using the definition of the derivative, find $f'(0)+f'(2)$.

At 0 $\dfrac{f(h)-f(0)}{h}=\dfrac{h|h|}{h}=|h|\to0$ from both sides, so $f'(0)=0$ โ€” unlike $|x|$, this function is differentiable at $0$.
At 2 Near $x=2$, $f(x)=x^2$, so $f'(2)=2\cdot2=4$.
Answer $0+4=\mathbf{4}$.
๐Ÿ”ฌ Researchโ€“Teaching NexusMini project ยท about 60 minutes

Research question. How fast was India's population growing in a given year โ€” and how well does a difference quotient from census data estimate that instantaneous growth rate?

Why it matters. Governments plan schools, hospitals and water supply from growth rates, not just totals. Census data come only every ten years, so planners estimate the rate at a particular year from differences โ€” exactly the difference quotient $\tfrac{P(t+h)-P(t)}{h}$ โ€” and they must decide which $h$ to use and whether to look forward, backward or both ways.

What to do.

  1. Collect India's census populations for $1971$, $1981$, $1991$, $2001$ and $2011$ (in crores) from an official or encyclopaedia source.
  2. Plot $P$ against the year and sketch by eye the tangent at $1991$. Estimate its slope (crores per year).
  3. Compute the forward difference quotient $\tfrac{P(2001)-P(1991)}{10}$, the backward one $\tfrac{P(1991)-P(1981)}{10}$, and their average, the centred quotient $\tfrac{P(2001)-P(1981)}{20}$. Which is closest to your tangent slope?
  4. Fit a quadratic $P(t)=a+bt+ct^2$ (with $t$ = years after $1971$) to three of the points in a spreadsheet, differentiate it, and compare $P'(20)$ with your quotients.
  5. Express each estimate as a percentage growth rate per year, $\tfrac{P'(t)}{P(t)}\times100\%$.

Tools: A spreadsheet or graphing app, census data

Hand in: Your table of data and quotients, the graph with the tangent at $1991$, the fitted model and a paragraph comparing the forward, backward and centred estimates.

Going further. For a quadratic model, show algebraically that the centred quotient $\tfrac{P(t+h)-P(t-h)}{2h}$ equals $P'(t)$ exactly, for every $h$. Why does that make the centred quotient the better estimate for real data?

Instructor note: approximate census figures are $54.8$, $68.3$, $84.6$, $102.9$ and $121.1$ crore. Forward quotient at $1991$: $1.83$ crore/year; backward: $1.63$; centred: $1.73$ โ€” about $2.0\%$ per year of $84.6$ crore. For $P=a+bt+ct^2$, $\tfrac{P(t+h)-P(t-h)}{2h}=b+2ct=P'(t)$ exactly, which students can verify in two lines.

๐Ÿงช Self-check quiz

0/5 correct

Q1. The derivative of $f$ at $a$ is defined as

Q2. From first principles, the derivative of $x^2$ at $x=5$ is

Q3. With $x$ in radians, $\dfrac{d}{dx}(\cos x)$ is

Q4. The slope of the tangent to $y=\dfrac1x$ at the point $(1,1)$ is

Q5. Which statement about $f(x)=|x|$ at $x=0$ is correct?

โ† 12.2 Limits of Polynomial, Rational and Trigonometric Functions12.4 Algebra of Derivatives โ†’