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What a function approaches, not what it equals
Say you want one number for how steep a curve is — not on average over some stretch, but exactly at one point. The slope between two points is easy: rise over run. Squeeze the second point down until it sits on the first, and the run becomes zero — rise over zero, undefined, nothing to divide. Most of us would stop there and call the question unanswerable.
Try it anyway on $f(x) = (x^2-1)/(x-1)$ at x=1. Direct substitution gives $0/0$ — no answer. But simplify first: for any x other than 1, $(x^2-1)/(x-1) = x+1$. As x gets closer to 1 from either side, $x+1$ gets closer to 2. The function is not even defined at x=1, yet the value it is heading toward is perfectly clear.
That heading-toward value is a limit. A limit asks what value a function APPROACHES as its input approaches some point — not what the function equals AT that point, which may be undefined, different, or the same value. The derivative, this chapter’s real subject, is built entirely on that distinction:
$f'(x) = \lim_{h \to 0} (f(x+h) - f(x))/h$
evaluates a ratio undefined exactly at h=0. It is the limit, not a direct substitution, that gives it a value there.
Read every claim in this chapter for what the function approaches, never for what it equals.
When does a limit exist?
Sometimes “the function approaches something” has two different answers depending on which side you arrive from. Picture a function that jumps at x=a: coming in from the left it climbs toward 3, but coming in from the right it climbs toward 5. Which one is the limit at a?
Neither, on its own. A limit at x=a exists only if two one-sided limits agree. The LEFT-HAND LIMIT ($\lim_{x \to a^-} f(x)$, x approaching a from below) must exist. The RIGHT-HAND LIMIT ($\lim_{x \to a^+} f(x)$, x approaching a from above) must exist too, and equal the first. When both hold, that common value is the limit.
If the two one-sided limits differ, or either fails to exist, the two-sided limit does not exist at that point. This holds regardless of what $f(a)$ itself is — even if f is not defined there at all.
Check both sides before writing down an answer: a single substitution never tells you whether the two sides agree.
One test decides it: do the left-hand and right-hand limits agree?
Compute $f'(1)$ for $f(x) = x^2$, from the definition
- $f'(1) = \lim_{h \to 0} [(1+h)^2 - 1]/h$
Form the difference quotient directly from the definition. - $(1+h)^2 - 1 = 2h + h^2$
Expand the numerator. - $[2h + h^2]/h = 2+h$
Divide by h — legal for any $h \neq 0$, exactly the regime the limit lives in. - $\lim_{h \to 0} (2+h) = 2$
Let h go to zero. - slope $= 2$ at $(1,1)$
That 2 is the SLOPE of the tangent line to $y=x^2$ at (1,1) — the limiting slope of the secant through (1,1) and $(1+h, (1+h)^2)$ as h shrinks to zero. Every derivative-at-a-point computation in this chapter repeats this same shape: form the difference quotient, simplify away the h, then let h go to zero. The number a derivative returns is always a slope first, a rate of change wearing whatever units the problem gives it.
Building limits from simpler limits — the algebra of limits
Working every limit from the raw definition — forming a difference quotient, simplifying, letting h shrink to zero — would make even a simple polynomial limit painfully slow. Four laws let you skip that, PROVIDED the pieces you are combining already have limits of their own.
Given $\lim_{x\to a} f(x) = l$ and $\lim_{x\to a} g(x) = m$, both already existing, four laws follow.
- SUM: $\lim_{x\to a} [f(x)+g(x)] = l+m$.
- DIFFERENCE: $\lim_{x\to a} [f(x)-g(x)] = l-m$.
- PRODUCT: $\lim_{x\to a} [f(x) \cdot g(x)] = l \cdot m$, which covers the scalar case $\lim_{x\to a} [c \cdot f(x)] = c \cdot l$ for constant c.
- QUOTIENT: $\lim_{x\to a} [f(x)/g(x)] = l/m$, provided $m \neq 0$.
Each law needs BOTH limits to already exist — the laws build a new limit out of two known ones; they do not create a limit where none exists. Miss that proviso and a law gets applied where it does not hold.
Four laws, one condition on every single one of them: the pieces must already have limits before you can combine them.
Evaluate $\lim_{x\to 2} [3x^2 + 5x - 1]$ by the algebra of limits
- $\lim_{x\to 2} [3x^2 + 5x - 1] = \lim_{x\to 2} 3x^2 + \lim_{x\to 2} 5x - \lim_{x\to 2} 1$
Split it into three pieces by the sum and difference laws — legal only because each piece is already known to have a limit. - $\lim_{x\to 2} 3x^2 = 3 (\lim_{x\to 2} x)^2 = 3(2^2) = 12$
Apply the scalar and product laws. - $\lim_{x\to 2} 5x = 5(2) = 10$
Apply the scalar law again; the constant term stays 1. - $12 + 10 - 1 = 21$
Add the pieces back — every step used one of the four algebra-of-limits laws, never a new rule, never a fresh appeal to the definition.
Limits of polynomial and rational functions
Every algebra-of-limits law from the last section has one job: build a new limit from ones already known to exist. Polynomials and rational functions are exactly where that job pays off immediately.
Take any polynomial f. It is built entirely from sums, products, and scalar multiples of x — nothing else. Apply the algebra-of-limits laws to each of those operations in turn, all the way down. The result is always $\lim_{x\to a} f(x) = f(a)$ — direct substitution works, every time, for every polynomial.
A rational function $f(x)/g(x)$ extends the same result one step further, through the quotient law: $\lim_{x\to a} f(x)/g(x) = f(a)/g(a)$, PROVIDED $g(a) \neq 0$. That proviso is not fine print — it is the quotient law’s own condition from the last section, carried forward untouched. The moment $g(a)=0$, the quotient law no longer applies, and direct substitution is not merely inconvenient; it is undefined.
Direct substitution is not a shortcut you reach for — it is what the algebra of limits already guarantees, right up until a denominator vanishes.
A denominator vanishing does not always mean the limit fails — it can mean the algebra hasn’t been simplified far enough yet.
Take a rational limit $\lim_{x\to a} f(x)/g(x)$ where BOTH $f(a)=0$ and $g(a)=0$ — the 0/0 indeterminate form. Direct substitution is undefined here, but the limit can still exist. Since f and g both vanish at x=a, the factor theorem guarantees that $(x-a)$ divides both of them exactly. Factor it out of numerator and denominator, cancel the shared $(x-a)$, and substitute into what remains.
This exact technique proves a genuinely general result. For any positive integer n,
$\lim_{x\to a} (x^n - a^n)/(x-a) = n a^{n-1} \cdot$
The factorisation $x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + \ldots + a^{n-1})$ cancels the $(x-a)$ directly, leaving n terms that each equal $a^{n-1}$ once $x=a$ is substituted — n copies of $a^{n-1}$, added together.
The same 0/0 shape sits behind average speed itself. Distance over time becomes exactly 0/0 the instant the time interval shrinks to zero — which is the whole reason a derivative needs a limit rather than plain division.
A 0/0 answer is not a dead end — it is a signal to factor first, substitute second.
Evaluate $\lim_{x\to 1} (x^3-1)/(x^2-1)$
- at $x=1$: $0/0$
Direct substitution gives $0/0$ on both counts — factor first. - $x^3-1 = (x-1)(x^2+x+1)$; $x^2-1 = (x-1)(x+1)$
Factor numerator and denominator. - $\lim_{x\to 1} (x^2+x+1)/(x+1)$
Cancel the common $(x-1)$. - $(1+1+1)/(1+1) = 3/2$
Now direct substitution works, because the denominator no longer vanishes at x=1.
Evaluate the same limit by the standard form
- $((x^3-1)/(x-1)) / ((x^2-1)/(x-1))$
Divide top and bottom by the same $x-1$. - $\lim_{x\to 1} (x^3-1)/(x-1) = 3 \times 1^2 = 3$
The standard form $\lim_{x\to a} (x^n-a^n)/(x-a) = n a^{n-1}$, with $n=3$, $a=1$. - $\lim_{x\to 1} (x^2-1)/(x-1) = 2 \times 1^1 = 2$
The same form again, $n=2$. - $3/2$
Same answer as factoring — two routes, one limit.
Limits of trigonometric functions
Polynomials and rationals fold into limits already known by direct substitution or factoring. Trigonometric functions near zero do neither — they need one new geometric fact, stated once and reused everywhere.
With x measured in RADIANS (not degrees — the geometric proof ahead depends on it), three standard limits hold.
- $\lim_{x\to 0} \sin(x)/x = 1$.
- $\lim_{x\to 0} \tan(x)/x = 1$ as well, and it follows directly from the first: $\tan(x)/x = [\sin(x)/x] \cdot [1/\cos(x)]$, and $\cos(0)=1$.
- $\lim_{x\to 0} (1-\cos(x))/x = 0$.
None of these three come from direct substitution — each one starts as $0/0$ at x=0 — and none come from the algebra-of-limits laws alone. They need a geometric argument, which the next section works through.
Three limits, one geometric proof behind all of them — stated here so the proof itself has something to aim at.
Why does $\sin(x)/x$ actually approach 1? Not by algebra — by comparing three areas on the unit circle.
For $0 < x < \pi/2$, picture the unit circle with angle x. Comparing the areas of two triangles and the circular sector trapped between them gives $\sin(x) < x < \tan(x)$. Divide through by $\sin(x)$, which is positive in this range, and take reciprocals — reversing both inequalities: $\cos(x) < \sin(x)/x < 1$.
As $x \to 0$, $\cos(x) \to 1$. $\sin(x)/x$ is sandwiched between $\cos(x)$ and 1, both closing in on 1 from either side, so $\sin(x)/x$ has nowhere else to go — it must approach 1 too. This is the SANDWICH theorem: trap an unknown quantity between two known ones heading to the same place, and it is forced there as well.
The same argument, run for $x \to 0^-$, gives the identical result — $\sin(x)/x$ is an even function of x, so the negative side mirrors the positive side exactly.
Three quantities, one squeeze — the limit has no room to be anything but 1.
Evaluate $\lim_{x\to 0} \sin(3x)/x$
- $\lim_{x\to 0} \sin(3x)/x$
The standard result is $\sin(x)/x \to 1$, but here the argument of sine is $3x$, not x — match them first. - $3 \cdot [\sin(3x)/(3x)]$
Rewrite by multiplying and dividing by 3, so the same quantity $3x$ sits in both the sine and the denominator. - as $x \to 0$, $3x \to 0$ too
The bracket approaches 1 by the standard trig limit. - $3 \cdot 1 = 3$
Whenever the argument inside sine does not match the denominator, force them to match first — multiply and divide by whatever it takes.
Building derivatives from simpler derivatives — the algebra of derivatives
The limit definition $f'(x) = \lim_{h\to 0} [f(x+h)-f(x)]/h$ computes any derivative directly. Re-deriving it from scratch for every function every time is exactly as slow as it sounds. Four rules, each proved once from that same definition, let a derivative be built from simpler derivatives instead.
Given $u=f(x)$ and $v=g(x)$, both differentiable.
- SUM/DIFFERENCE: $(u \pm v)' = u' \pm v'$.
- PRODUCT: $(u \cdot v)' = u' \cdot v + u \cdot v'$.
- QUOTIENT: $(u/v)' = (u'v - u v')/v^2$, provided $v \neq 0$.
- SCALAR MULTIPLE: $(c \cdot u)' = c \cdot u'$ for constant c.
The product and quotient rules are the two to watch. They are NOT $(u \cdot v)'=u' \cdot v'$ and NOT $(u/v)'=u'/v'$ — multiplying or dividing the two derivatives directly is the single most common algebra slip this toolkit invites. Each rule is proved by applying the algebra-of-limits laws to the difference quotient of the combined function — the same machine that built the limit toolkit builds this one too.
Four rules to reach for, and exactly one trap: never multiply or divide the derivatives directly for a product or a quotient.
Differentiate $f(x) = x^2 \sin(x)$
- $u = x^2$, $v = \sin(x)$
It is a product of two functions, so the product rule applies directly rather than expanding anything by hand. - $u' = 2x$
The same pattern the general $(x^n-a^n)/(x-a) \to n a^{n-1}$ result already produces, named properly as the power rule next section. - $v' = \cos(x)$
The derivative of sine, proved from first principles two sections ahead — taken here as a preview, the way a rule is often used once before it is formally derived. - $f'(x) = u'v + u v' = 2x \sin(x) + x^2 \cos(x)$
The product rule gives two terms, one from differentiating each factor while holding the other fixed — never $u'v'$ alone.
Differentiate $f(x) = x/(x+1)$
- $u = x$, $v = x+1$
Both are linear, so both derivatives come straight from the definition. - $[(x+h)-x]/h = h/h = 1$, so $u' = 1$
The difference quotient for x itself — even this obvious derivative comes from the same definition as everything else in this chapter. - $v' = 1$
The same reasoning gives $v'=1$ for $x+1$ — the added constant cancels in the difference quotient. - $f'(x) = (u'v-u v')/v^2 = [(x+1) - x] / (x+1)^2 = 1/(x+1)^2$
Apply the quotient rule.
Derivatives of polynomial and trigonometric functions
$d/d x (x^n) = n x^{n-1}$, for any positive integer n — this is the general pattern the last two worked examples already leaned on, now stated and derived properly.
The derivative definition gives $\lim_{h\to 0} [(x+h)^n-x^n]/h$. Substitute $y = x+h$, so $y-a=h$ with $a=x$, and $y \to a$ exactly as $h \to 0$. The limit becomes $\lim_{y\to a} (y^n-a^n)/(y-a)$ — the exact general result already proved: $n a^{n-1}$. With $a=x$, that is $n x^{n-1}$.
One substitution turns the derivative definition into a limit already known — the power rule is not a new fact, it is the old one renamed.
$d/d x \sin(x) = \cos(x)$ — the two worked examples that already used this fact get their proof here, from the definition directly.
From the limit definition,
$d/d x \sin(x) = \lim_{h\to 0} [\sin(x+h)-\sin(x)]/h \cdot$
The sum-to-product identity $\sin(A)-\sin(B) = 2 \cos((A+B)/2) \sin((A-B)/2)$, with $A=x+h$ and $B=x$, turns the numerator into $2 \cos(x+h/2) \sin(h/2)$. Dividing by h and regrouping: $\cos(x+h/2) \cdot [\sin(h/2)/(h/2)]$.
As $h \to 0$: $\cos(x+h/2) \to \cos(x)$, and $\sin(h/2)/(h/2) \to 1$ by the standard trig limit proved earlier this chapter. What remains is exactly $\cos(x)$.
The identical method, using $\cos(A)-\cos(B) = -2 \sin((A+B)/2) \sin((A-B)/2)$ in place of the sine identity, gives $d/d x \cos(x) = -\sin(x)$.
AC current and sound waves are both modelled as sinusoids — computing how either changes moment-to-moment is exactly this limit, run on a real signal instead of an abstract x.
The same three-step shape from every derivative in this chapter — form the difference quotient, simplify, let h shrink to zero — proves this one too.
Differentiate $f(x) = x^3 + \cos(x)$
- $f(x) = x^3 + \cos(x)$
It is a sum of two terms, so the sum rule applies — differentiate each term separately, then add. - $d/d x(x^3) = 3x^2$
Apply the power rule. - $d/d x(\cos(x)) = -\sin(x)$
Apply the first-principles result just proved. - $f'(x) = 3x^2 - \sin(x)$
Add the two terms — two rules from two different sections, used together on one function; that combination, not any single formula, is the actual skill.
One operation, many rules
Every result in this chapter, from the very first worked example to the last, reduces to one operation: evaluating
$\lim_{h\to 0} [f(x+h)-f(x)]/h \cdot$
Sometimes that limit is worked out once, in general, and handed over as a reusable RULE — the power rule, the algebra of derivatives, sin(x) turning into cos(x). Sometimes it is applied directly to one specific function, with no rule to reach for. Knowing which situation you are in — reach for a rule, or fall back to first principles — is the chapter’s real skill. No single formula in it matters as much as that judgment.
Economics gives this idea its own working name. The derivative of a cost or revenue function is called MARGINAL cost or marginal revenue — the field’s native vocabulary for exactly the same rate-of-change idea this chapter builds from scratch.
One operation, many disguises — everything in this chapter is that same limit, computed once and reused, or computed fresh.
Two traps this chapter sets
TRAP: if $\lim_{x\to a} f(x)$ exists, it feels like f must be defined at a, and its value there must equal the limit.
REALITY: a limit describes only the APPROACH — values of f NEAR a, taken as x gets arbitrarily close, and the definition never once looks at x=a itself. $f(a)$ can be undefined. It can be defined to a completely different value. Either way, the limit at a is entirely unaffected — this chapter’s very first example, $f(x)=(x^2-1)/(x-1)$, is undefined exactly at the point its limit describes.
This does not mean $f(a)$ is irrelevant everywhere — it means the limit and the function’s value are two separate questions, and this chapter has only ever answered the first one.
Check what f approaches. Never assume it tells you what f equals there.
TRAP: dy/dx looks like a fraction — a top part and a bottom part — so it feels safe to split it apart and recombine the two pieces like any ordinary fraction.
REALITY: dy/dx is ONE symbol for a single limit, $f'(x) = \lim_{h\to 0} [f(x+h)-f(x)]/h$. Later algebra sometimes treats it like a fraction — the chain rule mnemonic is the most common place this shows up — but that is a convenient resemblance, not the definition. Nothing in $\lim_{h\to 0} [f(x+h)-f(x)]/h$ splits into two independently meaningful parts called dy and dx.
The notation looks like a fraction. The definition underneath it isn’t one.
Practice set
- practice Evaluate $\lim_{x\to 3} (x^2-9)/(x-3)$.
- practice Evaluate $\lim_{x\to 0} (x^2+1)/(x+3)$.
- practice Does $\lim_{x\to 0} \left|x\right|/x$ exist? Justify using the left-hand and right-hand limits.
- practice Evaluate $\lim_{x\to 2} (x^3-8)/(x-2)$.
- practice Which of these equals $\lim_{x\to 4} (x^2-16)/(x-4)$?
- practice Evaluate $\lim_{x\to 2} (x^5-32)/(x-2)$.
- practice Evaluate $\lim_{x\to 3} (x^2-x-6)/(x-3)$.
- practice Let $f(x) = x+1$ for $x<2$ and $f(x) = 5$ for $x\geq 2$. Does $\lim_{x\to 2} f(x)$ exist? Justify using the left-hand and right-hand limits.
- practice Which of these equals $\lim_{x\to 5} (x^2-25)/(x-5)$?
- JEE $f(x) = (x^2-9)/(x-3)$ for $x \neq 3$, and $f(3) = 10$. Find $\lim_{x\to 3} f(x)$.
- JEE Using $f'(a) = \lim_{h\to 0} [f(a+h)-f(a)]/h$, find $f'(3)$ for $f(x) = 2x^2-5$.
- JEE $g(x) = 3$ for every $x \neq 2$, and $g(2) = 7$. Find $\lim_{x\to 2} [g(x) + 5]$.
- JEE $f(x) = \left|x-2\right|/(x-2)$. Does $\lim_{x\to 2} f(x)$ exist?
- JEE $f(x) = x+1$ for $x<1$, $f(1)=3$, $f(x)=x^2$ for $x\geq 1$. Find $\lim_{x\to 1} f(x)$, or state that it does not exist.
- JEE $f(x) = [x]$ (the greatest integer $\leq x$). Find $\lim_{x\to 2} [x]$, or state that it does not exist.
- JEE Given $\lim_{x\to 1} f(x) = 3$ and $\lim_{x\to 1} g(x) = -2$, find $\lim_{x\to 1} [2f(x) - 3g(x)]$.
- JEE Given $\lim_{x\to 2} f(x) = 5$ and $\lim_{x\to 2} g(x) = 3$, find $\lim_{x\to 2} [f(x) \cdot g(x) - 2f(x)]$.
- JEE Given $\lim_{x\to 0} f(x) = 2$, find $\lim_{x\to 0} [f(x)]^3$.
- JEE Evaluate $\lim_{x\to -1} [2x^3 - 3x^2 + 4]$.
- JEE Evaluate $\lim_{x\to 1} [(x^2+3)(x-2)]$.
- JEE Evaluate $\lim_{x\to -2} [x^3 + 3x^2 - x + 5]$.
- JEE Evaluate $\lim_{x\to -1} (x^2+2x+5)/(x+3)$.
- JEE Evaluate $\lim_{x\to 4} (x-1)/(x-4)$, or state that it does not exist.
- JEE Evaluate $\lim_{x\to -2} (x^3+1)/(x+5)$.
- JEE Evaluate $\lim_{x\to 2} (x^4-16)/(x-2)$.
- JEE Evaluate $\lim_{x\to -1} (x^3+1)/(x+1)$.
- JEE Evaluate $\lim_{x\to 3} (x^5-243)/(x-3)$.
- JEE Evaluate $\lim_{x\to 2} (x^2-x-2)/(x-2)$.
- JEE Evaluate $\lim_{x\to -3} (x^2+2x-3)/(x+3)$.
- JEE Evaluate $\lim_{x\to 2} (x^3-8)/(x^2-4)$.
- JEE $f(x) = (x^2-1)/(x-1)$ for $x \neq 1$, and $f(1)=100$. Find $\lim_{x\to 1} f(x)$.
- JEE $g(x) = 2x+1$ for every $x \neq 3$, and $g$ is left undefined at $x=3$. Find $\lim_{x\to 3} g(x)$.
- JEE $h(x) = (x^3-8)/(x-2)$ for $x \neq 2$, and $h(2)=0$. Find $\lim_{x\to 2} h(x)$.
Answers
- $6$
- $1/3$
- DNE — LHL $=-1$, RHL $=1$.
- $12$
- $8$
- $80$
- $5$
- DNE — LHL $=3$, RHL $=5$.
- $10$
- $6$
- $12$
- $8$
- does not exist — LHL $=-1$, RHL $=1$, and they disagree
- does not exist — LHL $=2$, RHL $=1$
- does not exist — LHL $=1$, RHL $=2$
- $12$
- $5$
- $8$
- $-1$
- $-4$
- $11$
- $2$
- does not exist
- $-7/3$
- $32$
- $3$
- $405$
- $3$
- $-4$
- $3$
- $2$
- $7$
- $12$
- practice Evaluate $\lim_{x\to 0} \sin(5x)/x$.
- practice Evaluate $\lim_{x\to 0} \tan(x)/x$.
- practice Which of these equals $\lim_{x\to 0} (1-\cos(x))/x$?
- JEE 2014 Evaluate $\lim_{x\to 0} \sin(\pi \cos(x)^2)/x^2$.JEE Main 2014 (Offline)
- practice Evaluate $\lim_{x\to 0} \sin(7x)/\sin(3x)$.
- practice Evaluate $\lim_{x\to 0} (1-\cos(2x))/(2x)$.
- JEE Evaluate $\lim_{x\to 0} \sin(7x)/x$.
- JEE Evaluate $\lim_{x\to 0} \tan(4x)/(3x)$.
- JEE Evaluate $\lim_{x\to 0} (1-\cos(6x))/x$.
- JEE Evaluate $\lim_{x\to 0} x \cdot \cot(x)$.
- JEE Evaluate $\lim_{x\to 0} \sin(2x)/\sin(3x)$.
- JEE Evaluate $\lim_{x\to 0} (1-\cos(x))/x^2$.
- JEE Evaluate $\lim_{x\to 0} \tan(5x)/x$.
- JEE Evaluate $\lim_{x\to 0} [\sin(4x) - \sin(2x)]/x$.
- JEE Evaluate $\lim_{x\to 0} \sin^2(3x)/x^2$.
Answers
- $5$
- $1$
- $0$
- $\pi$
- $7/3$
- $0$
- $7$
- $4/3$
- $0$
- $1$
- $2/3$
- $1/2$
- $5$
- $2$
- $9$
- practice Differentiate $f(x) = x^7 - 3x^4 + 2x$.
- practice Differentiate $f(x) = x \sin(x)$ using the product rule.
- practice Differentiate $f(x) = (x+1)/(x-1)$ using the quotient rule.
- practice Which of these equals $d/d x \cos(x)$?
- JEE 2019 If $\lim_{x\to 1} (x^2-a x+b)/(x-1) = 5$, find $a+b$.JEE Main 2019, 10th April (Evening Slot)
- practice Differentiate $f(x) = x^4 + 3 \sin(x) - 5$.
- practice Differentiate $f(x) = (x^2+1) \cos(x)$ using the product rule.
- practice Differentiate $f(x) = \sin(x)/x$ using the quotient rule.
- practice Using the limit definition directly, find $\lim_{h\to 0} [(x+h)^3 - x^3]/h$.
- JEE Using first principles, find $f'(3)$ for $f(x) = x^2$.
- JEE Using first principles, find $f'(1)$ for $f(x) = x^2 + 2x$.
- JEE Using first principles, find $f'(2)$ for $f(x) = x^2 - 4x$.
- JEE Given $f(1)=1$, $f'(1)=2$, $g(1)=2$, $g'(1)=1$, find $d/d x[2f(x) - 5g(x)]$ at $x=1$.
- JEE Given $f(1)=1$, $f'(1)=2$, $g(1)=2$, $g'(1)=1$, find $d/d x[f(x) \cdot g(x)]$ at $x=1$.
- JEE Given $f(1)=1$, $f'(1)=2$, $g(1)=2$, $g'(1)=1$, find $d/d x[f(x)/g(x)]$ at $x=1$.
- JEE Differentiate $f(x) = x \cdot \sin(x)$ using the product rule. Find $f'(\pi)$.
- JEE Differentiate $f(x) = x^3 \cdot \cos(x)$ using the product rule. Find $f'(\pi/2)$.
- JEE Differentiate $f(x) = \sin(x) \cdot \cos(x)$ using the product rule. Find $f'(\pi/4)$.
- JEE Differentiate $f(x) = x^2/(x+2)$ using the quotient rule. Find $f'(1)$.
- JEE Differentiate $f(x) = (2x+1)/(x-3)$ using the quotient rule. Find $f'(1)$.
- JEE Differentiate $f(x) = (x-1)/(x+1)$ using the quotient rule. Find $f'(2)$.
- JEE Find $d/d x(x^10)$ at $x=2$.
- JEE Find $d/d x(x^6)$ at $x=-1$.
- JEE Find $d/d x(x^5)$ at $x=2$.
- JEE Find $d/d x[\sin(x)]$ at $x=\pi/3$.
- JEE Find $d/d x[\cos(x)]$ at $x=\pi/6$.
- JEE Find $d/d x[\sin(x) - \cos(x)]$ at $x=\pi/4$.
- JEE Find $d/d x[x^4 - \sin(x)]$ at $x=\pi$.
- JEE Find $d/d x[2x^2 + 3\cos(x)]$ at $x=\pi/2$.
- JEE Find $d/d x[x^3 - 5\sin(x) + 2]$ at $x=0$.
- JEE For $y=x^3$, compare the TRUE derivative $d y/d x$ at $x=2$ (from the definition) against the finite ratio $[f(3)-f(2)]/(3-2)$ a student computes and calls '$d y/d x$'. What is the true $d y/d x$ at $x=2$?
- JEE For $y=x^2$, compare the TRUE derivative $d y/d x$ at $x=1$ against a finite ratio a student computes and calls '$d y/d x$'. What is the true $d y/d x$ at $x=1$?
- JEE For $y=x^2$, compare the TRUE derivative $d y/d x$ at $x=5$ against a finite ratio a student computes and calls '$d y/d x$'. What is the true $d y/d x$ at $x=5$?
- JEE Differentiate $x^2 \cdot \sin(x) + 3x$, choosing the right combination of rules. Find $f'(\pi)$.
- JEE Differentiate $x^3 \cdot \sin(x) - 2x^2$, choosing the right combination of rules. Find $f'(\pi)$.
- JEE Differentiate $(x+1)^2 \cdot \sin(x)$, choosing the right combination of rules (expand the square first, no chain rule needed). Find $f'(0)$.
Answers
- $7x^6 - 12x^3 + 2$
- $\sin(x) + x \cos(x)$
- $-2/(x-1)^2$
- $-\sin(x)$
- $-7$
- $4x^3 + 3 \cos(x)$
- $2x \cos(x) - (x^2+1) \sin(x)$
- $(x \cos(x) - \sin(x))/x^2$
- $3x^2$
- $6$
- $4$
- $0$
- $-1$
- $5$
- $3/4$
- $-\pi$
- $-\pi^3/8$
- $0$
- $5/9$
- $-7/4$
- $2/9$
- $5120$
- $-6$
- $80$
- $1/2$
- $-1/2$
- $\sqrt{2}$
- $1+4\pi^3$
- $2\pi-3$
- $-5$
- $12$
- $2$
- $10$
- $3-\pi^2$
- $-\pi(\pi^2+4)$
- $1$
Chapter-end problems
- board-easy Evaluate $\lim_{x\to 2} (3x^2 - 4x + 1)$.
- board-easy Evaluate $\lim_{x\to 1} (2x+3)/(x^2+1)$.
- board-easy Evaluate $\lim_{x\to -7} (x^2-49)/(x+7)$.
- board-easy Evaluate $\lim_{x\to 1} (x^6-1)/(x-1)$.
- board-easy Evaluate $\lim_{x\to 0} \sin(4x)/x$.
- board-easy Which of these equals $\lim_{x\to 0} (1-\cos(3x))/x$?
- board-easy Let $f(x) = x+3$ for $x<2$ and $f(x) = x^2-1$ for $x\geq 2$. Does $\lim_{x\to 2} f(x)$ exist? Justify using the left-hand and right-hand limits.
- board-easy Differentiate $f(x) = 5x^4 - 2x^3 + 7$ using the power rule.
- board-easy Differentiate $f(x) = 4x^3 + 2\sin(x) - 3$.
- board-easy Using first principles, find $f'(2)$ for $f(x) = x^2 + 3$.
- board-standard Evaluate $\lim_{x\to 4} (\sqrt{x}-2)/(x-4)$.
- board-standard Evaluate $\lim_{x\to 0} (\sqrt{1+x}-\sqrt{1-x})/x$.
- board-standard Given $\lim_{x\to 1} f(x) = 2$ and $\lim_{x\to 1} g(x) = 5$, find $\lim_{x\to 1} [f(x)^2 - 3g(x)]/[g(x)-1]$.
- board-standard Evaluate $\lim_{x\to 0} \sin(5x)/\tan(3x)$.
- board-standard Let $f(x) = x^2-1$ for $x\leq 1$ and $f(x) = 2x-2$ for $x>1$. Does $\lim_{x\to 1} f(x)$ exist? Justify using the left-hand and right-hand limits.
- board-standard Differentiate $f(x) = (3x^2+1)(2x-5)$ using the product rule.
- board-standard Differentiate $f(x) = (2x-1)/(3x+4)$ using the quotient rule.
- board-standard Differentiate $f(x) = x^2 \tan(x)$ using the product rule.
- board-standard Differentiate $f(x) = \cos(x)/(x+1)$ using the quotient rule.
- board-standard Which of these equals $d/d x[\tan(x)]$ at $x=\pi/4$?
- board-standard Using first principles, find $f'(1)$ for $f(x) = x^2 + 3x$.
- board-standard Which of these equals $\lim_{x\to 0} \tan(2x)/\sin(4x)$?
- JEE If $\lim_{x\to 3} (x^2+a x+b)/(x-3) = 7$, find $a+b$.
- JEE Let $f(x) = 2x+3$ for $x<1$ and $f(x) = k x^2$ for $x\geq 1$. Find the value of $k$ for which $\lim_{x\to 1} f(x)$ exists.
- JEE Which of these equals $\lim_{x\to 0} (1-\cos(4x))/x^2$?
- JEE Using first principles, find $f'(x)$ for $f(x) = \sqrt{x}$.
- JEE Differentiate $f(x) = x^2 (x+1)(2x-3)$, choosing the right combination of rules.
- JEE Find the point on the curve $y = x^2-4x+3$ where the tangent is parallel to the x-axis.
- JEE Which of these equals $\lim_{x\to 0} [\sqrt{1+x}-1]/x$?
- JEE Given $f(1)=3$, $f'(1)=-2$, $g(1)=4$, $g'(1)=1$, find $d/d x[f(x)/g(x) + 3f(x)]$ at $x=1$.
Answers
- $5$
- $5/2$
- $-14$
- $6$
- $4$
- $0$
- does not exist — LHL $=5$, RHL $=3$
- $20x^3 - 6x^2$
- $12x^2 + 2\cos(x)$
- $4$
- $1/4$
- $1$
- $-11/4$
- $5/3$
- $0$
- $18x^2-30x+2$
- $11/(3x+4)^2$
- $2x \tan(x) + x^2 \sec^2(x)$
- $-[(x+1)\sin(x)+\cos(x)]/(x+1)^2$
- $2$
- $5$
- $1/2$
- $-11$
- $5$
- $8$
- $1/(2\sqrt{x})$
- $8x^3-3x^2-6x$
- $(2,-1)$
- $1/2$
- $-107/16$