Tool for calculating the domain of differentiability of a function f(x) ie., all values of x for which the derivative function f'(x) yields a value.
Domain of Derivative of a Function - dCode
Tag(s) : Functions
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The domain of differentiability of a function is the set of values of $ x $ for which the function is defined and differentiable. It corresponds to the domain of definition of its derivative, denoted by $ D_{f'} $.
Determining the domain of differentiability of a function consists of identifying the values of $ x $ for which the function is differentiable. This domain must be a subset of the function's domain of definition.
The domain of differentiability can be determined in 2 steps:
Step 1: Determine the values of $ x $ for which the function is differentiable, taking into account its domain of definition, differentiation rules and any points where it is not differentiable.
Step 2: Calculate the derivative of the function on this domain.
Example: $ f(x) = \ln(x) $ is defined and differentiable on $ \mathbb{R}^{\*+} = ] 0 ; +\infty [ $. Its derivative is $ f'(x) = \frac{1}{x} $, defined on $ D_{f'} = \mathbb{R}^{\*+} = ] 0 ; +\infty [ $.
A rational function of the form $ f(x) = \frac{P(x)}{Q(x)} $, where $ P $ and $ Q $ are polynomials and $ Q $ is not the zero polynomial, is differentiable wherever $ Q(x) \neq 0 $. Its domain of differentiability is therefore equal to its domain of definition.
Indeed, its derivative is given by $ f'(x) = \frac{P'(x)Q(x) - P(x)Q'(x)}{Q(x)^2} $. This expression is defined for every value of $ x $ such that $ Q(x) \neq 0 $. Even if the fraction can be simplified algebraically, the domain of definition of the original function remains determined by the original expression for $ Q(x) $.
A function that is differentiable at a point must be continuous at that point. Furthermore, a function cannot be differentiable at a point where it is not defined.
Thus, the domain of differentiability of a function is a subset of its domain of definition. Every function that is differentiable on a set is also continuous on that set, but the converse is not always true: a function can be continuous without being differentiable at certain points.
Example: The function $ f(x) = |x| $ is defined and continuous on $ \mathbb{R} $, but it is not differentiable at $ x = 0 $. Its domain of differentiability is therefore $ \mathbb{R}^{\*} = ] -\infty ; 0 [ \cup ] 0 ; +\infty [ $.
To determine the points where a function is not differentiable, first identify its domain of definition, then look for points where the derivative does not exist. Several situations are possible:
— The function is not defined at the point under consideration.
— The function is discontinuous.
— The function has a corner point where the left-hand and right-hand derivatives differ.
— The function has a vertical tangent, with a difference quotient whose magnitude tends to infinity.
Example: The function $ f(x) = |x| $ is not differentiable at $ x = 0 $, because its left-hand derivative is $ -1 $ and its right-hand derivative is $ 1 $.
No, a continuous function is not necessarily differentiable. Differentiability implies continuity, but continuity does not imply differentiability. A function can be continuous at a point without having a derivative there, particularly if it has a corner point or a vertical tangent.
Example: The function $ f(x) = |x| $ is continuous on $ \mathbb{R} $, but is not differentiable at $ x = 0 $, because its left-hand and right-hand derivatives differ.
To determine the domain of differentiability of a composite function of the form $ f(x) = g(h(x)) $, first determine the values of $ x $ for which $ h(x) $ is defined and belongs to the domain of definition of $ g $. Then check that $ h $ is differentiable at $ x $ and that $ g $ is differentiable at $ h(x) $.
When these conditions are met, the composite function is differentiable and its derivative is given by the chain rule: $ f'(x) = g'(h(x)) \times h'(x) $. The domain of differentiability therefore includes the values satisfying these conditions, but it may be larger if the composite function remains differentiable at certain points where one of these sufficient conditions is not met.
Example: For $ f(x) = \sqrt{x^2 + 1} $, the inner function $ h(x) = x^2 + 1 $ is defined and differentiable on $ \mathbb{R} $, and the square root function is differentiable for every strictly positive argument. Since $ x^2 + 1 > 0 $ for every real number $ x $, the composite function is differentiable on $ \mathbb{R} $.
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