AP Calculus AB/BC

Part 2 · differentiation

The differentiation rules

Power, product, quotient. Each is a shortcut for a limit you would otherwise compute by hand, and each has a failure mode worth naming.

By the end of this chapter you can

  • Differentiate powers, including negative and fractional exponents
  • Apply the product and quotient rules correctly
  • Recognise when a rule does not apply
  • Differentiate the standard functions from memory

Chapter 2.1 computed a derivative from the definition. Doing that every time would be unbearable, so this chapter is the set of shortcuts — each one a theorem proved once from the limit, then used forever.

The rules are worth knowing cold. The failure modes are worth knowing better, because that is where the marks go.

The power rule

ddx xn=nxn−1\frac{d}{dx}\,x^n = n x^{n-1}

  • ddxx5=5x4\frac{d}{d x} x^{5} = 5 x^{4}
  • ddxx=12x\frac{d}{d x} \sqrt{x} = \frac{1}{2 \sqrt{x}}
  • ddx1x=−1x2\frac{d}{d x} \frac{1}{x} = - \frac{1}{x^{2}}
  • ddxx=1\frac{d}{d x} x = 1
checked by the build

It holds for every real nn, not just positive integers, and the second and third claims above are the ones students forget. Rewrite before differentiating:

  • x=x1/2\sqrt{x} = x^{1/2}, so its derivative is 12x−1/2\tfrac{1}{2}x^{-1/2}.
  • 1x=x−1\tfrac{1}{x} = x^{-1}, so its derivative is −x−2-x^{-2}.
  • ddx2x=2xlog⁡(2)\frac{d}{d x} 2^{x} = 2^{x} \log{\left(2 \right)}
  • ddxex=ex\frac{d}{d x} e^{x} = e^{x}
checked by the build

exe^x is the function that is its own derivative, and that is what makes ee special rather than any aesthetic property.

Linearity

Derivatives pass through sums and constant multiples untouched:

ddx(af(x)+bg(x))=af′(x)+bg′(x)\frac{d}{dx}\big(a f(x) + b g(x)\big) = a f'(x) + b g'(x)

  • ddx(3x2+5x−7)=6x+5\frac{d}{d x} \left(3 x^{2} + 5 x - 7\right) = 6 x + 5
  • ddx(−2ex+4sin⁡(x))=−2ex+4cos⁡(x)\frac{d}{d x} \left(- 2 e^{x} + 4 \sin{\left(x \right)}\right) = - 2 e^{x} + 4 \cos{\left(x \right)}
checked by the build

This is the rule that makes polynomials trivial, and it is the reason the next two rules come as a surprise — multiplication and division do not behave this way.

The product rule

The tempting guess is that the derivative of a product is the product of the derivatives. It is not, and one example settles it: if it were, then ddx(x⋅x)\tfrac{d}{dx}(x \cdot x) would be 1⋅1=11 \cdot 1 = 1, when it is plainly 2x2x.

ddx(fg)=f′g+fg′\frac{d}{dx}\big(f g\big) = f' g + f g'

  • ddxxex=xex+ex\frac{d}{d x} x e^{x} = x e^{x} + e^{x}
  • ddxx2sin⁡(x)=x2cos⁡(x)+2xsin⁡(x)\frac{d}{d x} x^{2} \sin{\left(x \right)} = x^{2} \cos{\left(x \right)} + 2 x \sin{\left(x \right)}
  • ddxx2=2x\frac{d}{d x} x^{2} = 2 x
checked by the build

The last line is the counterexample, verified: the product rule gives 1⋅x+x⋅1=2x1 \cdot x + x \cdot 1 = 2x, which agrees with the power rule. Any proposed rule has to survive that test.

Why two terms. Each factor changes, and each change contributes. Hold gg still and vary ff: you get f′gf'g. Hold ff still and vary gg: you get fg′fg'. The real change is both at once, and the term where both change is proportional to h2h^2, which the limit discards.

The quotient rule

ddx(fg)=f′g−fg′g2\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f' g - f g'}{g^2}

  • ddxsin⁡(x)x=cos⁡(x)x−sin⁡(x)x2\frac{d}{d x} \frac{\sin{\left(x \right)}}{x} = \frac{\cos{\left(x \right)}}{x} - \frac{\sin{\left(x \right)}}{x^{2}}
  • ddxx+1x−1=−2(x−1)2\frac{d}{d x} \frac{x + 1}{x - 1} = - \frac{2}{\left(x - 1\right)^{2}}
  • ddxtan⁡(x)=1cos⁡2(x)\frac{d}{d x} \tan{\left(x \right)} = \frac{1}{\cos^{2}{\left(x \right)}}
checked by the build

Two things go wrong with this one, reliably:

The order matters. The numerator is f′g−fg′f'g - fg', not fg′−f′gfg' - f'g. Unlike the product rule, swapping the terms flips the sign of every answer. A memory hook that survives pressure: the term starting with the derivative of the top comes first.

It is often not worth using. x2+1x\tfrac{x^2+1}{x} is easier rewritten as x+x−1x + x^{-1} and differentiated term by term. Reach for the quotient rule when the denominator genuinely resists splitting.

  • x2+1x=x+1x\frac{x^{2} + 1}{x} = x + \frac{1}{x}
  • ddx(x+1x)=1−1x2\frac{d}{d x} \left(x + \frac{1}{x}\right) = 1 - \frac{1}{x^{2}}
checked by the build

The standard derivatives

These must be immediate, because everything else is built on them.

ddxsin⁡x=cos⁡xddxcos⁡x=−sin⁡xddxtan⁡x=sec⁡2x\frac{d}{dx}\sin x = \cos x \qquad \frac{d}{dx}\cos x = -\sin x \qquad \frac{d}{dx}\tan x = \sec^2 x

ddxex=exddxln⁡x=1x\frac{d}{dx}e^x = e^x \qquad \frac{d}{dx}\ln x = \frac{1}{x}

  • ddxsin⁡(x)=cos⁡(x)\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}
  • ddxcos⁡(x)=−sin⁡(x)\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}
  • ddxlog⁡(x)=1x\frac{d}{d x} \log{\left(x \right)} = \frac{1}{x}
  • ddx(sin⁡2(x)+cos⁡2(x))=0\frac{d}{d x} \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right) = 0
checked by the build

That last claim is a pleasant consistency check: sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 is constant, so its derivative must be zero — and applying the rules gives 2sin⁡xcos⁡x−2cos⁡xsin⁡x2\sin x\cos x - 2\cos x \sin x, which is zero. When a derivative you compute contradicts something you know, one of them is wrong, and this is a cheap way to find out which.