6. DERIVATIVES OF ELEMENTARY FUNCTIONS

The mentioned below formulae for derivatives of elementary functions should be remembered. They remain valid, provided that both sides are meaningful.

c ' = 0 for any number c ∈ R,
(xα) ' = α · x^α-1, for every non-zero number α,
(e^x) ' = e^x,
(a^x) ' = a^x lna, for every number a > 0
(sinx) ' = cosx,
(cosx) ' = -sinx,
(tgx) ' = 1/cos² x,
(ctgx) ' = -1/sin² x.

The expression "both sides are meaningful" means for instance:

Theorem (the derivative of a compound function)

If functions f and g are differentiable at points x₀ and f (x₀), respectively, then

(g · f) ' (x₀) = g ' (f (x₀)) · f ' (x₀).

Example

Making use of the formulae for derivatives of elementary functions and the last theorem, we will find the derivatives of the following functions:

h (x) = e^tgx
Using notation of the last theorem we have f (x) = tgx and g (x) = e^x. Hence

(g · f) ' (x) = [ g (f (x)) ] ' = (e^{tg x}) ' = e^{tg x} · (1 / cos²x )
y = sin (cosx^1/3)
This function is a composition of three functions: f (x) = x^1/3, g (x) = cosx and h (x) = sinx, so

y ' (x) = (h · g · f) ' = h ' ( (g · f) (x) ) · (g · f) ' (x) = h ' ( (g · f) (x) ) · g ' (f (x)) · f ' (x) =

= cos (cosx^1/3) · (-sinx^1/3) · (1/3) x^-2/3.
This function is a composition of three functions:

Hence

y ' (x) = (h · g · f) ' = h ' ( (g · f) (x) ) · (g · f) ' (x) = h ' ( (g · f) (x) ) · g ' (f (x)) · f ' (x) =

By the last theorem formulae for derivatives of functions f (ax) and f (ax+b) where a,b ∈ R are of the form:

[ ( f (ax) ) ] ' = a f ' (ax)

[ ( f (ax + b) ) ] ' = a f ' (ax + b)

Example

Theorem

Let f be a continuous and single-valued function. If f is differentiable at a point x₀ and its derivative at this point is different from zero, then a function f ^-1 is differentiable at a point y₀ = f (x₀) and:

(f^-1) ' (y₀) = 1 / f ' (x₀).

Using the last theorem one derives the following formulae:

Theorem (the logarithmic derivative)

If a function f is differentiable on an open interval P, f(x) > 0 for x ∈ P, then

f ' (x) = f (x) · ( ln f (x) ) '.

Proof

Computing the derivative of lnf (x), making use of the theorem on the derivative of a compound function, we have

(ln f (x) ) ' = ( 1/f (x) ) · f ' (x).

The last theorem is most frequently used for evaluation of derivatives of functions of the form f (x) ^g(x).

Example

We will find the derivative of the function f (x) = (sinx)^x.

We have:

f ' (x) = (sinx)^x [ ln (sinx)^x ] ' = (sinx)^x [ x ln (sinx) ] ' = (sinx)^x [ ln( sinx) + x · cosx · (1/sinx) ].

We have made use of the product rule.

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