Unizor- Derivatives - Compound Functions

Notes to a video lecture on http://www.unizor.com

Derivative Properties -
Compound Functions

Our purpose is to express the derivative of a compound function (a function of a function) in terms of derivatives of its components.

Assume that the real functionf(x) is defined anddifferentiable (that is, its derivative exist) on certain interval with a derivative f ^I(x).
Assume further that the real function g(x) is defined anddifferentiable on certain interval with a derivative g ^I(x) and its values are falling within the domain of function f(x).
Then we can talk about acompound functionh(x)=f(g(x)).

In other words, this compound function can be represented as
h(x)=f(y), where y=g(x).

Let's determine the derivative of this compound function.
The increment of functionh(x)=f(g(x)) is
Δh(x) = h(x+Δx)−h(x) =
= f(g(x+Δx)) − f(g(x))

Using the definition of the increment of a function
g(x+Δx) = g(x) + Δg(x),
it would look like
Δh(x) = h(x+Δx)−h(x) =
= f(g(x)+Δg(x)) − f(g(x))

Recalling the representationy=g(x), we can write this as
Δh(x) = f(y+Δy) − f(y)
where y=g(x).

Next operations to find a derivative are: dividing the function increment Δh(x) by an increment of an argument Δxand going to a limit as Δx→0.

Δh(x)/Δx =
= [f(y+Δy) − f(y)]/Δx =
= Δf(y)/Δx
where y=g(x).

To transform this into expressions that would lead us to separation of functions combined in this compound function, multiply and divide this by Δy.
Δh(x)/Δx =
= [Δf(y)/Δy]·[Δy/Δx]

Notice that if Δx→0 then Δy→0 since function y=g(x) is differentiable and, since Δy→0, Δf(y)→0 since function f(x) is differentiable.

Therefore, if Δx→0,
[Δf(y)/Δy] → df(y)/dy
where y=g(x) and
[Δy/Δx] → dy/dx=dg(x)/dx

Finally, we came to an expression for a derivative of a compound function.
Δh(x)/Δx → [df(y)/dy]·[dy/dx]
In other words,
dh(x)/dx = [df(y)/dy]·[dy/dx]
where y=g(x).

In terms of original functions,
df(g(x))/dx =
= [df(y)/dy]·[dg(x)/dx]
where y=g(x).
This is a major formula of differentiation of a compound function.

Example 1
f(x)=cos(x); g(x)=x²
f(g(x))=cos(x²)
dcos(x²)/dx =
= [dcos(y)/dy]·[d(x²)/dx] =
= [−sin(y)]·[2x]
where y=x².
Therefore,
dcos(x²)/dx = −2·x·sin(x²)

Example 2
f(x)=1/x(x); g(x)=cos(x)
f(g(x))=1/cos(x)=sec(x)
[sec(x)]^I =
=[1/cos(x)]^I =
=[1/y]^I·[cos(x)]^I =
=[−1/y²]·[−sin(x)]
where y=cos(x).
Therefore,
[sec(x)]^I = sin(x)/cos²(x)

Example 3
f(x)=e^x; g(x)=sin(x)
f(g(x))=e^sin(x)
[e^sin(x)]^I =
=[e^y]^I·[sin(x)]^I =
=[e^y]·[cos(x)]
where y=sin(x).
Therefore,
[e^sin(x)]^I = e^sin(x)·cos(x)

Example 4
f(x)=x²; g(x)=sin(x)
f(g(x))=(sin(x))²=sin²(x)
[sin²(x)]^I =
=[y²]^I·[sin(x)]^I =
=[2y]·[cos(x)]
where y=sin(x).
Therefore,
[sin²(x)]^I = 2·sin(x)·cos(x)