Intuition What this page is forThe parent note gave you the six formulas and how they're born. This page is the drill floor : we hunt down every kind of problem the topic can throw at you and solve one of each — signs, domains, chain-rule traps, degenerate inputs, blow-up points, a word problem, and an exam twist. By the end you should never meet a case you haven't already seen defeated.
Keep these two "engine" formulas in view; almost everything below is one of them plus the Chain Rule :
d x d arctan x = 1 + x 2 1 , d x d arcsin x = 1 − x 2 1 .
Common mistake The four traps we keep pointing at (named so you don't have to jump back)Throughout this page we refer to four classic errors. Here they are, defined once:
Trap A — dropping the chain-rule factor. Writing d x d arctan ( 3 x ) = 1 + 9 x 2 1 1 + 9 x 2 3 arctan ( x ) x u ′ Trap B — wrong sign on a "co-" function. Thinking d x d arccos x = + 1 − x 2 1 decreases , so its slope must be negative .Trap C — dropping the ∣ x ∣ Writing x x 2 − 1 1 both branches, so its derivative is positive everywhere; the ∣ x ∣ Trap D — ignoring the domain. Trying to compute d x d arcsin x x = 2 1 − x 2
Before solving anything, let's list every distinct kind of situation this topic contains. Each row is a "cell". Every cell gets covered by at least one worked example below.
Cell
What makes it different
Example that hits it
C1 Plain formula, positive inputjust quote the derivative, no chain
Ex 1
C2 Chain rule, linear insideextra constant factor appears
Ex 2
C3 Chain rule, nonlinear insideinside squared/shifted, watch the domain
Ex 3
C4 A "co-" function (sign trap)must carry the minus sign
Ex 4
C5 arcsec/arccsc with $x
$
C6 Negative input / sign of slopecheck slope stays positive/negative as theory says
Ex 6
C7 Limiting / blow-up value (x → ± 1 derivative → ∞
Ex 7
C8 Degenerate / out-of-domain inputrecognise "no answer exists"
Ex 8
C9 Word problem (rate of change)inverse trig models a real angle
Ex 9
C10 Exam twist (identity + product)combine two rules, simplify cleverly
Ex 10
C11 arccot (the sixth function)last of the six, its own minus sign
Ex 11
Let me plant one picture first so every "slope" claim below has something to point at.
Intuition Read the pictureThe red curve is arcsin x only lives between x = − 1 x = 1 (dashed walls), it is always rising (slope always positive), and it gets vertical at the two edges — that is the 1 − x 2 → 0 C6, C7, C8 are all just "which part of this red curve am I standing on?"
Worked example Differentiatef ( x ) = arctan x x = 1 Forecast: guess before reading — is the slope at x = 1 1
Step 1. Quote the engine formula. Why this step? No inside function other than x Chain Rule correction is owed.
f ′ ( x ) = 1 + x 2 1 .
Step 2. Plug in x = 1 Why this step? "Slope at a point" means evaluate the derivative there.
f ′ ( 1 ) = 1 + 1 2 1 = 2 1 .
Verify: arctan x π /2 less than 1 . We got 2 1 < 1
Worked example Differentiateg ( x ) = arcsin ( 5 x ) Forecast: will the answer be 1 − 25 x 2 1
Step 1. Identify inside u = 5 x Why this step? The formula d x d arcsin u = 1 − u 2 u ′ u u ′ = 5
Step 2. Apply the formula, replacing x u = 5 x Why this step? The 1 − □ 2 whole inside thing squared.
g ′ ( x ) = 1 − ( 5 x ) 2 1 ⋅ 5 = 1 − 25 x 2 5 .
Verify: the domain is now ∣5 x ∣ < 1 ∣ x ∣ < 5 1 narrower window than plain arcsin 5 x ± 1 5 Trap A ). ✓
Worked example Differentiateh ( x ) = arctan ( x 2 + 1 ) Forecast: the inside is never 0 1 + ( inside ) 2
Step 1. Inside u = x 2 + 1 u ′ = 2 x Why this step? We must differentiate the inside separately — that's the whole Chain Rule .
Step 2. Slot into d x d arctan u = 1 + u 2 u ′ Why this step? arctan h ′ ( x ) = 1 + ( x 2 + 1 ) 2 2 x .
Step 3. Optionally expand ( x 2 + 1 ) 2 = x 4 + 2 x 2 + 1 1 + x 4 + 2 x 2 + 1 = x 4 + 2 x 2 + 2 Why this step? Cleaner for later integration checks.
h ′ ( x ) = x 4 + 2 x 2 + 2 2 x .
Verify: the denominator x 4 + 2 x 2 + 2 ≥ 2 > 0 all real x — matching that arctan x = 0 h ′ = 0 arctan ( x 2 + 1 ) x = 0
Worked example Differentiatep ( x ) = arccos ( 2 x ) x = 0 Forecast: positive or negative slope? Remember arccos goes downhill .
Step 1. Recall d x d arccos u = − 1 − u 2 u ′ minus . Why this step? arccos is a "co-" function; it decreases (bigger ratio → Trap B ).
Step 2. Inside u = 2 x u ′ = 2 Why this step? The chain rule supplies the factor u ′ = 2 p ′ ( x ) = − 1 − 4 x 2 2 .
Step 3. Evaluate at x = 0 Why this step? "Slope at x = 0 x = 0 p ′ p ′ ( 0 ) = − 1 2 = − 2.
Verify: slope is negative , exactly as arccos's downhill shape demands. If you got + 2
Worked example Differentiateq ( x ) = arcsec ( x ) x = 2 x = − 2 arccsc ( x ) Forecast: arcsec increases on both branches — so should the slope have the same sign at x = 2 x = − 2 opposite sign?
Step 1. Quote d x d arcsec x = ∣ x ∣ x 2 − 1 1 Why the ∣ x ∣ Without it the algebra gives x x 2 − 1 1 negative on the left branch — but arcsec rises there too, so the derivative must stay positive . The ∣ x ∣ Trap C ).
Step 2. At x = 2 Why this step? Direct evaluation of the derivative at the point.
q ′ ( 2 ) = ∣2∣ 4 − 1 1 = 2 3 1 = 6 3 .
Step 3. At x = − 2 Why this step? To test the "same sign on both branches" claim, we must actually compute the left branch.
q ′ ( − 2 ) = ∣ − 2∣ ( − 2 ) 2 − 1 1 = 2 3 1 = 6 3 .
Step 4. The companion. Why this step? The cell is "arcsec /arccsc ", so we owe the csc case too. arccsc is the "co-" partner of arcsec, so by the parent's pattern its derivative is the negative of arcsec's:
d x d arccsc x = − ∣ x ∣ x 2 − 1 1 . x = 2 − 6 3 x = − 2 − 6 3 negative slope on both branches, because arccsc decreases everywhere.
Verify: arcsec came out equal and positive on both branches (+ 6 3 ≈ 0.2887 ∣ x ∣ x equal and negative (− 6 3
Worked example Find the slope ofarcsin x x = − 2 1 Forecast: the red curve in the figure rises everywhere — so the slope at a negative x positive , right?
Step 1. Write down the derivative. Why this step? "Slope of arcsin x f ′ ( x ) = 1 − x 2 1 .
Step 2. Substitute x = − 2 1 Why this step? "Slope at x = − 2 1 1 − x 2 1 − x 2 = 1 − 4 1 = 4 3 ⟹ f ′ ( − 2 1 ) = 3/4 1 = 3 2 = 3 2 3 .
Verify: the answer is positive even though x < 0 x squared in the formula, the sign of x 3 2 3 ≈ 1.1547 > 1 4 5 ∘
Worked example What happens tod x d arcsin x x → 1 − Forecast: finite number, or infinite?
Step 1. Look at the denominator 1 − x 2 x → 1 − Why this step? The whole behaviour of the slope is controlled by that square root shrinking.
1 − x 2 → 1 − 1 = 0 + ⟹ 1 − x 2 → 0 + .
Step 2. Divide a fixed numerator by that vanishing denominator. Why this step? A constant 1 0 + lim x → 1 − 1 − x 2 1 = + ∞.
Verify (geometry): in the picture below the red curve becomes vertical at the walls x = ± 1 ± 1
Intuition Why this figure mattersThis second picture is the proof by eyeball of Ex 7. The black curve is the same arcsin x x = 1 vertical . When a graph goes vertical, "rise over run" has a run heading to zero — that's the 1 − x 2 → 0 → + ∞ no slope at all — there's no curve out there to be tangent to.
d x d arcsin x x = 2 Forecast: try to guess an answer — then check whether the point even exists.
Step 1. Test the domain first. Why this step? arcsin ∣ x ∣ ≤ 1 Inverse Functions and their Domains ).
Step 2. Substitute x = 2 1 − x 2 Why this step? The formula's square root needs 1 − x 2 ≥ 0 1 − x 2 = 1 − 4 = − 3 < 0 ⟹ 1 − x 2 = − 3 is not real .
Conclusion: the derivative does not exist at x = 2 This is the correct answer , not a number (Trap D ).
Verify: the figures' red curve simply stops at x = 1 x = 2
Worked example A person stands10 m top edge is at height x θ = arctan ( 10 x ) θ x = 10 m Forecast: at x = 10 4 5 ∘
Step 1. Differentiate θ x u = 10 x u ′ = 10 1 Why this step? "How fast does θ d x d θ d x d θ = 1 + ( 10 x ) 2 1 ⋅ 10 1 = 10 1 ⋅ 1 + 100 x 2 1 = 100 + x 2 10 . Why the algebra? Multiply top and bottom by 100
Step 2. Evaluate at x = 10 Why this step? The question asks for the rate at that particular height , so we substitute x = 10 d x d θ x = 10 = 100 + 100 10 = 200 10 = 20 1 = 0.05 rad/m .
Verify (units + sense): units are radians per metre ✓. Value 0.05 ≈ 2. 9 ∘ 4 5 ∘
Worked example Differentiater ( x ) = x arctan x − 2 1 ln ( 1 + x 2 ) Forecast: exam questions like this usually collapse to something tiny. Guess what r ′ ( x )
Step 1. Split r product rule . Why this step? x arctan x x x arctan x d x d ( uv ) = u ′ v + u v ′ u = x , v = arctan x d x d ( x arctan x ) = u ′ 1 ⋅ arctan x + x ⋅ v ′ 1 + x 2 1 = arctan x + 1 + x 2 x . Why v ′ = 1 + x 2 1 That's just the engine formula for arctan
Step 2. Differentiate the second piece − 2 1 ln ( 1 + x 2 ) logarithm–chain rule . Why this step? ln d x d ln ( w ) = w w ′ w = 1 + x 2 w ′ = 2 x d x d ( − 2 1 ln ( 1 + x 2 ) ) = − 2 1 ⋅ 1 + x 2 2 x = − 1 + x 2 x . Why the − 2 1 It's a constant multiplier, so it just rides along unchanged.
Step 3. Add the two derivatives and watch the cancellation. Why this step? r ′ 1 + x 2 x r ′ ( x ) = ( arctan x + 1 + x 2 x ) + ( − 1 + x 2 x ) = arctan x + 1 + x 2 x − 1 + x 2 x = arctan x .
Verify: the extra fractions annihilate, leaving the clean r ′ ( x ) = arctan x r ( x ) the antiderivative of arctan x Integration by Inverse Trig result read backwards. Sanity check at x = 0 r ′ ( 0 ) = arctan 0 = 0 0 + 0 − 0 = 0
Worked example Differentiates ( x ) = arccot ( 4 x ) x = 0 Forecast: arccot is the "co-" partner of arctan, so its base derivative carries a minus . Positive or negative slope at x = 0
Step 1. Recall the sixth engine formula d x d arccot x = − 1 + x 2 1 Why the minus? arccot decreases everywhere (bigger ratio → Trap B in disguise).
Step 2. Inside u = 4 x u ′ = 4 Why this step? The Chain Rule supplies the factor u ′ = 4 1 + □ 2 s ′ ( x ) = − 1 + ( 4 x ) 2 1 ⋅ 4 = − 1 + 16 x 2 4 .
Step 3. Evaluate at x = 0 Why this step? "Slope at x = 0 x = 0 s ′ s ′ ( 0 ) = − 1 + 0 4 = − 4.
Verify: the slope is negative (− 4 + 1 + x 2 1 arccot x
Recall Which cell was which?Plain formula ::: Ex 1 (C1)
Chain rule with a linear inside gives an extra constant factor ::: Ex 2 (C2)
Nonlinear inside, domain stays all reals for arctan ::: Ex 3 (C3)
"co-" function needs the minus sign ::: Ex 4 (C4)
arcsec has ∣ x ∣ x → + ∞ x → 1 − d x d θ = 100 + x 2 10 0.05 x = 10 d x d ( x arctan x − 2 1 ln ( 1 + x 2 ) ) = arctan x d x d arccot ( 4 x ) = − 1 + 16 x 2 4 − 4 x = 0
Mnemonic Three questions before you differentiate"Domain? Inside? Sign?" — (1) Is the input allowed (Trap D , C8)? (2) Is there an inside function needing the Chain Rule (Trap A , C2, C3, C11)? (3) Is it a "co-" function that owes a minus (Trap B , C4, C11)?
Multiply by inside derivative
