Differentiate the outer function at the inner, then multiply by the inner’s derivative. The engine behind almost every rule that follows.
Find the inner by the “last operation” test: whatever you would compute last plugging in a number is the outer; what’s left inside is \(u\). Peel outermost-first, one layer at a time.
trap
- The inner derivative \(u'\) is the most-dropped factor — \(\frac{d}{dx}\sin(x^2)=\cos(x^2)\cdot 2x\), not \(\cos(x^2)\).
- Evaluate \(f'\) at the inner value, not at \(x\): in a table, \(h'(1)=f'\!\bigl(g(1)\bigr)\,g'(1)\).
Every one is the same move — outer derivative \(\times\) inner derivative \(u'\). Memorize the shapes and the chain becomes automatic.
power
\(\frac{d}{dx}u^{n}=n\,u^{n-1}u'\)
exponential
\(\frac{d}{dx}e^{u}=e^{u}\,u'\)
natural log
\(\frac{d}{dx}\ln u=\dfrac{u'}{u}\)
general base
\(\frac{d}{dx}a^{u}=a^{u}\ln a\;u'\)
root
\(\frac{d}{dx}\sqrt{u}=\dfrac{u'}{2\sqrt{u}}\)
sine
\(\frac{d}{dx}\sin u=\cos u\;u'\)
When you can’t solve for \(y\): treat \(y\) as a hidden function of \(x\), so every \(y\)-term carries a \(\frac{dy}{dx}\).
The four moves
1 · differentiateboth sides w.r.t. \(x\)
2 · tagevery \(y\)-term gets a \(\frac{dy}{dx}\) (chain rule)
3 · collectall \(\frac{dy}{dx}\) on one side, factor
4 · solvedivide for \(\frac{dy}{dx}\)
trap
- Treating \(y^2\) as a constant — it differentiates to \(2y\,y'\), never \(2y\).
- \(xy\) needs the product rule: \(\frac{d}{dx}(xy)=y+x\,y'\).
Once you have \(\frac{dy}{dx}=\frac{N}{D}\), the special tangents are pure sign-reading.
Horizontal or vertical?
numerator \(=0\)horizontal tangent (slope \(0\))
denominator \(=0\)vertical tangent (slope undefined)
implicit 2nd derivative (circle)
trap
- Substitute \(y'\) back in after differentiating \(y'\) — then simplify with the original equation.
- Both conditions need the other side nonzero — \(\frac00\) is indeterminate, not a tangent.
Just differentiate again — and again. Notation stacks; patterns emerge.
Two patterns worth knowing: the trig cycle \(\sin\!\to\!\cos\!\to\!-\sin\!\to\!-\cos\) repeats every 4; and \(\frac{d^n}{dx^n}\frac1x=\dfrac{(-1)^n n!}{x^{n+1}}\) — alternating sign, factorial coefficient.
trap
- Differentiate fully, then evaluate — never plug in a number between derivatives.
- Watch the sign by parity: even-order derivatives of \(1/x\) and of \(\cos\) flip predictably.
Reflecting a graph across \(y=x\) swaps rise and run, so the inverse’s slope is the reciprocal — taken at the matched point.
Table method: to get \((f^{-1})'(c)\), find \(c\) in the output row, hop to its input \(a=f^{-1}(c)\), then reciprocate: \(1/f'(a)\).
trap
- Reciprocate at \(f^{-1}(x)\), not at \(x\). The single most common error — always solve for the matched input first.
- Reflection keeps the sign of the slope: a rising function has a rising inverse.
Six to know; co-functions are just the negatives. Chain rule: replace \(x\) with \(u\) and multiply by \(u'\).
arcsin
\(\dfrac{1}{\sqrt{1-x^2}}\)
arccos
\(-\dfrac{1}{\sqrt{1-x^2}}\)
arctan
\(\dfrac{1}{1+x^2}\)
arccot
\(-\dfrac{1}{1+x^2}\)
arcsec
\(\dfrac{1}{|x|\sqrt{x^2-1}}\)
arccsc
\(-\dfrac{1}{|x|\sqrt{x^2-1}}\)
trap
- \(\arctan\) has no radical — it’s \(\frac{1}{1+x^2}\); \(\arcsin\) has the \(\sqrt{\;}\).
- \(\operatorname{arcsec}\) keeps the \(|x|\) so the slope stays positive on both branches.
For a variable base and exponent, or a messy product/quotient: take \(\ln\) of both sides first, then differentiate implicitly.
Classic: \(y=x^{x}\Rightarrow \ln y=x\ln x\Rightarrow \frac{y'}{y}=\ln x+1\Rightarrow y'=x^{x}(\ln x+1)\).
trap
- Multiply back by \(y\) at the end — \(\frac{y'}{y}=\dots\) isn’t the answer until you clear the \(y\).
- Neither pure rule works on \(x^x\): the power rule needs a constant exponent, the exponential rule a constant base.
Differentiation is parsing. Name the outermost operation first — it picks the rule; recurse inward.
What is the outer operation?
sum / differencedifferentiate term by term
productproduct rule \(u'v+uv'\)
quotientquotient rule — or split / simplify first
compositionchain rule
tangled \(x\) & \(y\)implicit differentiation
trap
- Rewrite before you differentiate: log laws split \(\ln(x^4(x^2{+}1))\); \(\frac{x^2+1}{x}=x+\frac1x\); factor \(\frac{x^3-1}{x-1}\).
- Forcing one favorite rule onto everything — the outer operation, not habit, chooses.
The headline application: each derivative of position is the next kinematic quantity.
Speeding up when \(v\) and \(a\) share a sign; slowing down when they oppose. \(v=0\) does not force \(a=0\) — a particle can be momentarily still while accelerating.
trap
- Velocity is signed; speed is \(|v|\). “How fast” wants speed, direction wants the sign of \(v\).
- Turning points are where \(v=0\) and changes sign — check the sign, don’t assume.