Main Properties of Derivatives

Continuity

Theorem (Continuity of differentiable functions)
 
If a function \( f \) is differentiable at point \( a \) then it is continuous at \( a \).

Additive property

In this section we will prove that a sum of two differentiable functions is differentiable, and that a scalar multiple of a differentiable function is differentiable. We will also derive the properties for the derivative of a sum and the derivative of a scalar multiple of a function.

Theorem: Derivative of a scalar multiple
 
Assume that \( c\in\mathbb R \) and that \( f \colon\mathbb R\to\mathbb R \) is a differentiable function at point \( a \). Then \( g(x)=cf(x) \) is also differentiable at \( a \) and its derivative satisfies \( g^{\prime}(a)=cf^{\prime}(a) \).

Theorem: Derivative of a sum
 
If \( f \) and \( g \) are two differentiable functions at point \( a \), then so is \( h(x)=f(x)+g(x) \) and the derivative of \( h \) at \( a \) can be evaluated as: \[ h^{\prime}(a)=f^{\prime}(a)+g^{\prime}(a).\]

From the previous two theorems we can conclude that \[ (f-g)^{\prime}=\Big(f+(-1)\cdot g\Big)^{\prime}=f^{\prime}+(-1)\cdot g^{\prime}=f^{\prime}-g^{\prime}.\]

Product Rule

In this section we will derive the formula for the derivative of a product of two functions.
Theorem (Product rule)
 
If \( f \) and \( g \) are two differentiable functions at \( a\in\mathbb R \) that so is \( h=f\cdot g \) and the following formula holds: \[ h^{\prime}(a)=f^{\prime}(a)\cdot g(a)+f(a)\cdot g^{\prime}(a).\]

Example
 
Find the derivative of the function \( f(x)=x^2\cdot \cos x \).

Theorem (Derivative of a reciprocal)
 
If \( f \) is a function differentiable at \( a \) that satisfies \( f(a)\neq 0 \), then the function \( g(x)=\frac1{f(x)} \) is also differentiable at \( a \) and satisfies \[ g^{\prime}(a)=-\frac{f^{\prime}(a)}{f(a)^2}.\]

Using the previous two theorems we now establish the following result:

Theorem (Quotient rule)
 
If \( f \) and \( g \) are function differentiable at \( a \) such that \( g(a)\neq 0 \), then the function \( h(x)=\frac{f(x)}{g(x)} \) is differentiable at \( a \) and satisfies: \[ h^{\prime}(a)=\frac{f^{\prime}(a)\cdot g(a)-f(a)\cdot g^{\prime}(a)}{g(a)^2}.\]


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