###### Calculus I

Exercices

Applications

Marginal analysis

# Calculus I: Derivatives Proof of some derivative properties

In this chapter, we’re going to prove many of the various derivative properties, by using the definition:

### 1. Continuity of a differentiable function

If f(x) is differentiable at x = a then f(x) is continuous at x = a .

f(x) is differentiable at x = a . So
lim ((f(x) - f(a))/(x - a) = f'(a)
x → a

We have if x ≠ a :

lim (f(x) - f(a)) = lim [(f(x) - f(a))/(x - a)] . lim (x - a) =
x → a
= f'(a) x 0 = 0

We have also :

lim f(x) = lim (f(x) - f(a) + f(a) ) =
x → a
lim (f(x) - f(a)) + lim f(a) = 0 + f(a) = f(a)
x → a

lim f(x) = f(a) → f is continuoous at x = a
x → a

### 2. Sum and difference of two functions

(f(x) ± g(x))' = f'(x) ± g'(x)

We have :

lim [(f(x + h) + g(x + h)) - (f(x) + g(x))]/h = (f(x) + g(x))' =
h → 0

lim [f(x + h) - f(x) + g(x + h)) - g(x)]/h =
h → 0

lim (f(x + h) - f(x))/h + lim (g(x + h) - g(x))/h =
h → 0

f'(x) + g'(x))

(f(x) ± g(x))' = f'(x) ± g'(x))

### 3. Constant times a function

(c f(x))' = c f'(x)

We have :

lim (c f(x + h) - c f(x))/h = c lim (f(x + h) - f(x))/h = c f'(x)
h → 0

(c f(x))' = c f'(x)

### 4. Derivative of a constant

(c)' = 0

We have :

lim (c - c )/h = 0 = lim 1/h = 0
h → 0

(c)' = 0

### 5. Product rule

(f g)' = f' g + f g'

We have :

lim (f(x + h) g(x + h) - f(x) g(x))/h =
h → 0

hint : add and subtract f(x + h) g(x)

lim (f(x + h) g(x + h) - f(x) g(x) + f(x + h) g(x) - f(x + h) g(x))/h =
h → 0

lim (f(x + h) g(x + h) - f(x + h) g(x) - f(x) g(x) + f(x + h) g(x))/h =
h → 0

lim (f(x + h) (g(x + h) - g(x)) + g(x) (f(x + h) - f(x))) /h =
h → 0

lim (f(x + h) (g(x + h) - g(x)))/h + lim (g(x) (f(x + h) - f(x)))/h =
h → 0

f(x + 0) g'(x) + g(x) f'(x) = f(x) g'(x) + g(x) f'(x)

(f g)' = f' g + f g'

### 6. Quotient rule

(f/g)' = (f' g - f g')/g2

We have :

lim (f(x + h) / g(x + h) - f(x) / g(x))/ h =
h → 0

lim (1/h) (f(x + h) g(x) - f(x) g(x + h))/ g(x) g(x + h) =
h → 0

hint : add and subtract f(x) g(x)

lim (1/h) (f(x + h) g(x) - f(x) g(x + h) + f(x) g(x) - f(x) g(x) )/ g(x) g(x + h) =
h → 0

lim (1/h) (f(x + h) g(x) - f(x) g(x) + f(x) g(x) - f(x) g(x + h)) / g(x) g(x + h) =
h → 0

lim (1/h) (g(x)(f(x + h) - f(x)) - f(x)(g(x + h) - g(x))) / g(x) g(x + h) =
h → 0

lim (g(x)(f(x + h) - f(x))/h - f(x)(g(x + h) - g(x))/h / g(x) g(x + h)) =
h → 0

(g(x)f'(x) - f(x)g'(x)) / g(x) g(x + 0) =

(g(x)f'(x) - f(x)g'(x)) / (g(x))2

(f/g)' = (f'g) - fg')/g2

### 7. Power rule

(xn)' = n xn - 1

We can use three methods to prove this:

. Logarithmic differentiation,
. The derivative:
f'(a) = lim (f(x) - f(a))/(x - a), and
x → a
. The the binomial methode. Let's use this latter.

The binomial formula is :

(x + h)n = Σ (n, k) xn - k hk
Fom k = 0 to n .
Where (n, k) = n!/k!(n - k)!

Then

lim ((x + h)n - xn) / h = f'(x) =
h → 0

(Σ (n, k) xn - khk - xn) / h =
h → 0

(xn + n xn - 1 h + ... + - xn) / h =
h → 0

(n xn - 1 + ... + ) = n xn - 1

(xn)' = n xn - 1

### 8. Chain rule

(f(g(x)))' = f'(g(x)) . g'(x)

F(x) = fog(x) = f(g(x))
F'(x) = f'(g(x)) . g'(x)

lim (F(x + h) - F(x))/h = F'(x) =
h → 0

lim ( f(g(x + h)) - f(g(x)) )/h =
h → 0

We are assuming that the function g(x) is differentiable at the point x. That is the number g'(x) exists and is equal to the limit definition of the derivative:

lim (g(x + h) - g(x))/h = g'(x)
h → 0

We are assuming also that the function f(x) is differentiable at the point x. That is the number f'(x) exists and is equal to the limit definition of the derivative:

lim (f(y + k) - f(y))/k = f'(y)
k → 0

Let's rewrite f(g(x + h)) at h NEAR 0 :

g(x + h) = h g'(x) + g(x)
f(y + k) = k f'(y) + f(y)

And replace it in :

lim ( f(g(x + h)) - f(g(x)) )/h = F'(x)
h → 0

So

lim (f(g(x) + h g'(x)) - f(g(x)) )/h
h → 0

Let's use f(y + k) = k f'(y) + f(y), therefore:

f(g(x) + h g'(x)) = h g'(x) f'(g(x)) + f(g(x))

lim ( (h g'(x) f'(g(x)) + f(g(x))) - f(g(x)) )/h =
h → 0

lim ( h g'(x) f'(g(x)) )/h = g'(x) f'(g(x))
h → 0

(f(g(x)))' = f'(g(x)) . g'(x)

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