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Concept 05 / 12Limits, Continuity and Differentiability
Rules of Differentiation and Derivatives of Standard Functions
First-principles derivatives of x^n, sin x, e^x and ln x, the sum, product, quotient and chain rules, derivatives of trigonometric, inverse trigonometric, exponential and logarithmic functions, and simplifying inverse trigonometric expressions by substitution before differentiating.
Derivatives of the basic functions come from the definition and the standard limits.
dxdxn=nxn−1 (for any real n, where defined), from limh(x+h)n−xn.
dxdsinx=cosx: hsin(x+h)−sinx=h2cos(x+h/2)sin(h/2)→cosx.
dxdex=ex: hex+h−ex=exheh−1→ex. This is the property that makes e the natural base.
dxdlnx=x1 for x>0, and more generally dxdln∣x∣=x1 for x=0.
Rules.
Linearity: (af+bg)′=af′+bg′.
Product: (fg)′=f′g+fg′. For three factors, (fgh)′=f′gh+fg′h+fgh′.
Quotient: (gf)′=g2f′g−fg′ where g=0.
Chain: dxdf(g(x))=f′(g(x))⋅g′(x), or in Leibniz notation dxdy=dudy⋅dxdu. Differentiate the outer function at the inner value, then multiply by the derivative of the inner function; for nested compositions, keep multiplying.
Table of standard derivatives.
dxdcosx=−sinx; dxdtanx=sec2x; dxdcotx=−csc2x; dxdsecx=secxtanx; dxdcscx=−cscxcotx.
dxdax=axlna; dxdlogax=xlna1.
dxdsin−1x=1−x21 and dxdcos−1x=−1−x21 for ∣x∣<1.
dxdtan−1x=1+x21 and dxdcot−1x=−1+x21.
dxdsec−1x=∣x∣x2−11 for ∣x∣>1.
The inverse derivatives follow from the inverse function rule dydx=dy/dx1. For y=sin−1x, x=siny, so dydx=cosy=1−x2 (positive, since y∈[−2π,2π]).
Simplify inverse trigonometric expressions before differentiating. Substitutions such as x=tanθ, x=sinθ or x=cos2θ often collapse the expression. For example, y=tan−11−x22x: with x=tanθ, 1−x22x=tan2θ, so y=2θ=2tan−1x when ∣x∣<1, and dxdy=1+x22. The range restriction matters: for x>1, 2θ∈(2π,π) and y=2tan−1x−π, which has the same derivative, but for expressions like sin−11+x22x the derivative changes sign across ∣x∣=1: it is 1+x22 for ∣x∣<1 and −1+x22 for ∣x∣>1.
Derivative of a function with respect to another function: d(g(x))d(f(x))=g′(x)f′(x).
Key Takeaways:
- (fg)′=f′g+fg′; (gf)′=g2f′g−fg′; (f∘g)′=f′(g)g′.
- (ex)′=ex; (ax)′=axlna; (ln∣x∣)′=x1.
- (tanx)′=sec2x; (secx)′=secxtanx.
- (sin−1x)′=1−x21; (tan−1x)′=1+x21; (sec−1x)′=∣x∣x2−11.
- Simplify inverse trigonometric expressions with a substitution, then watch the interval.
- dgdf=g′(x)f′(x).
- Pitfall: forgetting the inner derivative in the chain rule, for example writing (sin3x)′=cos3x.
Brain-Teaser 1: Differentiate y=sin−1(2x1−x2) for 21<x<1.
Solution: Put x=sinθ with θ∈(4π,2π). Then 2x1−x2=2sinθcosθ=sin2θ, with 2θ∈(2π,π). The principal value sin−1(sin2θ) is π−2θ on that interval. So y=π−2sin−1x and dxdy=−1−x22. For ∣x∣<21 the answer would be +1−x22. The substitution works only if you track which branch of sin−1 applies.
Brain-Teaser 2: Find the derivative of sin2x with respect to ecosx.
Solution: dxd(sin2x)=2sinxcosx and dxd(ecosx)=−sinxecosx. The ratio is −sinxecosx2sinxcosx=−2cosxe−cosx. Differentiating both with respect to the common variable x and dividing is the whole method.
Brain-Teaser 3: Using first principles, show that dxdx=2x1 and explain why the derivative fails at x=0.
Solution: hx+h−x=h(x+h+x)(x+h)−x=x+h+x1→2x1 for x>0. At x=0 only the right-hand quotient exists: hh=h1→∞. So the graph has a vertical tangent at the origin and x is not differentiable there, although it is continuous. The formula 2x1 blowing up at 0 is the algebraic sign of the vertical tangent.
