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Mathematics · Ch 7 — Limits and Derivatives

Derivatives of Polynomial and Trigonometric Functions

10

Derivatives of Polynomial and Trigonometric Functions

With the algebra of derivatives (Section 9) in hand, only the derivatives

of the basic building blocks -- powers of xx, and sin⁡x\sin x, cos⁡x\cos x -- are still needed to

differentiate any polynomial or trigonometric expression built from sums, differences, products

and quotients of them. Each is derived here from first principles.

Derivative of a constant. For f(x)=cf(x)=c (constant), f′(x)=lim⁡h→0c−ch=lim⁡h→00=0f'(x)=\displaystyle\lim_{h\to0} \frac{c-c}{h}=\lim_{h\to0}0=0: a constant function has zero derivative everywhere, exactly as

expected for a function whose rate of change is always zero.

Theorem (power rule). For any positive integer nn,

ddx(xn)=n xn−1.\frac{d}{dx}\big(x^n\big) = n\,x^{n-1}.

Proof. By definition and the binomial theorem,

ddx(xn)=lim⁡h→0(x+h)n−xnh.\frac{d}{dx}\big(x^n\big) = \lim_{h\to0}\frac{(x+h)^n-x^n}{h}.

Expanding (x+h)n(x+h)^n by the binomial theorem,

(x+h)n=xn+(n1)xn−1h+(n2)xn−2h2+⋯+(nn)hn,(x+h)^n = x^n + \binom{n}{1}x^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots + \binom{n}{n}h^n,

so

(x+h)n−xn=nxn−1h+(n2)xn−2h2+⋯+hn=h[nxn−1+(n2)xn−2h+⋯+hn−1].(x+h)^n - x^n = n x^{n-1}h + \binom{n}{2}x^{n-2}h^2 + \cdots + h^n = h\left[n x^{n-1} + \binom{n}{2}x^{n-2}h + \cdots + h^{n-1}\right].

Dividing by hh and letting h→0h\to0, every term after the first carries a positive power of hh

and so vanishes, leaving only the first term:

ddx(xn)=nxn−1.■\frac{d}{dx}\big(x^n\big) = n x^{n-1}. \qquad\blacksquare

(The same formula also holds for negative integer and rational exponents nn, though the proof

differs slightly in each case and is not needed within this syllabus.) Combined with the sum and

scalar-multiple rules of Section 9, this immediately differentiates any polynomial term by term:

for p(x)=c0+c1x+c2x2+⋯+cnxnp(x)=c_0+c_1x+c_2x^2+\cdots+c_nx^n,

p′(x)=c1+2c2x+3c3x2+⋯+ncnxn−1.p'(x) = c_1 + 2c_2x + 3c_3x^2 + \cdots + nc_nx^{n-1}.

Theorem. ddx(sin⁡x)=cos⁡x.\dfrac{d}{dx}(\sin x) = \cos x.

Proof. By definition,

ddx(sin⁡x)=lim⁡h→0sin⁡(x+h)−sin⁡xh.\frac{d}{dx}(\sin x) = \lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}.

Use the sum-to-product identity sin⁡A−sin⁡B=2cos⁡ ⁣(A+B2)sin⁡ ⁣(A−B2)\sin A-\sin B = 2\cos\!\left(\dfrac{A+B}{2}\right) \sin\!\left(\dfrac{A-B}{2}\right) with A=x+h, B=xA=x+h,\ B=x, so A+B2=x+h2\dfrac{A+B}{2}=x+\dfrac{h}{2} and

A−B2=h2\dfrac{A-B}{2}=\dfrac{h}{2}:

sin⁡(x+h)−sin⁡x=2cos⁡ ⁣(x+h2)sin⁡ ⁣(h2).\sin(x+h)-\sin x = 2\cos\!\left(x+\frac{h}{2}\right)\sin\!\left(\frac{h}{2}\right).

So

sin⁡(x+h)−sin⁡xh=cos⁡ ⁣(x+h2)⋅sin⁡(h/2)h/2.\frac{\sin(x+h)-\sin x}{h} = \cos\!\left(x+\frac{h}{2}\right)\cdot\frac{\sin(h/2)}{h/2}.

As h→0h\to0: cos⁡ ⁣(x+h2)→cos⁡x\cos\!\left(x+\dfrac{h}{2}\right)\to\cos x (continuity of cosine, Section 4), and

sin⁡(h/2)h/2→1\dfrac{\sin(h/2)}{h/2}\to1 (the standard limit of Section 4, applied with h/2→0h/2\to0). Hence

ddx(sin⁡x)=cos⁡x⋅1=cos⁡x.■\frac{d}{dx}(\sin x) = \cos x\cdot1 = \cos x. \qquad\blacksquare

Theorem. ddx(cos⁡x)=−sin⁡x.\dfrac{d}{dx}(\cos x) = -\sin x.

Proof. By an identical argument, using cos⁡A−cos⁡B=−2sin⁡ ⁣(A+B2)sin⁡ ⁣(A−B2)\cos A-\cos B=-2\sin\!\left(\dfrac{A+B}{2}\right) \sin\!\left(\dfrac{A-B}{2}\right):

cos⁡(x+h)−cos⁡x=−2sin⁡ ⁣(x+h2)sin⁡ ⁣(h2)⟹cos⁡(x+h)−cos⁡xh=−sin⁡ ⁣(x+h2)⋅sin⁡(h/2)h/2 →h→0 −sin⁡x.■\cos(x+h)-\cos x = -2\sin\!\left(x+\frac{h}{2}\right)\sin\!\left(\frac{h}{2}\right) \quad\Longrightarrow\quad \frac{\cos(x+h)-\cos x}{h} = -\sin\!\left(x+\frac{h}{2}\right)\cdot\frac{\sin(h/2)}{h/2} \ \xrightarrow[h\to0]{}\ -\sin x. \qquad\blacksquare

Derivatives of the remaining trigonometric functions, obtained from these two using the

quotient rule of Section 9 (not re-derived from first principles): writing tan⁡x=sin⁡x/cos⁡x\tan x=\sin x/\cos x

with u=sin⁡x, u′=cos⁡x, v=cos⁡x, v′=−sin⁡xu=\sin x,\ u'=\cos x,\ v=\cos x,\ v'=-\sin x, …

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