The derivatives of sin(x) and cos(x) can be derived using the limit definition of the derivative. For sin(x),

d dx [sin(x)]	=	lim Δx→ 0  sin(x+Δx)-sin(x) Δx
	=	lim Δx→ 0  [sin(x)cos(Δx)+cos(x)sin(Δx)]-sin(x) Δx
	=	lim Δx→ 0  sin(x) cos(Δx) -1 Δx + cos(x) sin(Δx) Δx
	=	sin(x) lim Δx→ 0  cos(Δx) -1 Δx + cos(x) lim Δx→0  sin(Δx) Δx	These limits are not obvious! See your calculus text or otherwise convince yourself that they are correct.
	=	sin(x)(0) + cos(x)(1)
	=	cos(x).

The derivative of cos(x) is derived analogously. Then the remaining derivatives can be derived using the Quotient Rule, since all the other trigonometric functions are quotients involving sin(x) and cos(x).

Example

The derivative of tan(x²) is

d dx tan (x2)	=	sec2 (x2) · d dx (x2)
	=	2x sec2(x2)

by the Chain Rule.

Logarithmic Functions

By the definition of the natural logarithm, d/dx[lnx] = ¹/_x for x0. Also, d/dx[ln|x|] = ¹/_x for all x ≠ 0. To see this, suppose x < 0. Then ln|x| = ln(-x).

So

d dx [ln|x|]	=	d dx ln(-x)
	=	1  -x  · d dx (-x)
	=	1  -x  (-1)
	=	1 x

Example

By the Chain Rule, the derivative of ln(x³+5) is [1/( x³+5)]·[d/dx(x³+5)] = (3x²)/( x³+5).

Exponential Functions

There is an elegant way to show that d /dx [e^x] = e^x. We start with the identity ln(e^x) = x. Differentiating both sides,

d dx [ln(ex)]	=	d dx (x)
d dx [ln(ex)]	=	1
1 ex · d dx (ex)	=	1
d dx (ex)	=	ex.

Since e^x is never 0, this derivation holds for all x.

Example

The derivative of e^-3x+2 is e^-3x+2 · d/dx (-3x+2) = -3e^-3x+2.

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Key Concepts [index]

f(x)		f ′(x)
sin(x)		cos(x)
cos(x)		-sin(x)
tan(x)		sec2(x)
sec(x)		sec(x)tan(x)
csc(x)		-csc(x)cot(x)
cot(x)		-csc2(x)
ln(x)		1/x
ex		ex