# Derivative Calculator

This calculator **evaluates derivatives** using analytical differentiation. It will also
find **local minimum** and **maximum**, of the given function.
The calculator will try to simplify result as much as possible.

There are examples of valid and invalid expressions at the bottom of the page.

**Result**

### Help

1. For powers use ^. Example: x12= x^12 ; ex+2= e^(x+2)

2. For square root use "sqrt". Example: x+1−−−−−√= sqrt(x+1).

3. Supported constants: e, pi

4. Supported functions: sqrt, ln ( use 'ln' instead of 'log'), e (use 'e' instead of 'exp')

Trigonometric functions: sin cos tan cot sec csc

Inverse trigonometric functions: acos asin atan acot asec acsc

Hyperbolic functions: sinh, cosh, tanh, coth, sech, csch

## Examples of valid and invalid expressions

Function to integrate | Correct syntax is | Incorrect syntax is |

$$ (2x+1)^6 $$ | (2x+1)^6 | [2x+1]^6 |

$$ \frac{10x + 1}{x^2-4} $$ | (10x+1)/(x^2-4) | 10x+1/x^2-4 |

$$ \left(ln(x)\right)^2 $$ | ln(x)^2 | ln^2(x) |

$$ x ~ ln\left(\frac{x-1}{x+1}\right) $$ | x*ln((x-1)/(x+1)) | x*ln(x-1)/(x+1) |

The derivative of a function f at a point x, written ƒ ′(x), is given by:

– lim ƒ(x + ∆x) — ƒ(x)

ƒ′(x) = ∆s→0 ——————–

– ∆x

if this limit exists.

the derivative of a function corresponds to the slope of its tangent line at one specific point. The following illustration allows us to visualise the tangent line (in blue) of a given function at two distinct points. Note that the slope of the tangent line varies from one point to the next. The value of the derivative of a function therefore depends on the point in which we decide to evaluate it. By abuse of language, we often speak of the slope of the function instead of the slope of its tangent line.

## Derivatives of usual functions

Below you will find a list of the most important derivatives. Although these formulas can be formally proven, we will only state them here. We recommend you learn them by follows.

**The constant function**

Let ƒ(x) = k, where k is some real constant.

Then ƒ′(x) = (k)′ = 0

Example

(8)′ = 0

(—5)′ = 0

(0,2321)′ = 0

The identity function †(x) = x

Let ƒ(x) = x , the identity function of x. Then

ƒ′(x) = (x)′ = 1

• The rule mentioned above applies to all types of exponents (natural, whole, fractional). It is however essential that this exponent is constant. Another rule will need to be studied for exponential functions (of type ax ).

• The identity function is a particular case of the functions of form

• x^{n} (with n = 1) and follows the same derivation rule : (x)′ = (x1)′ = 1 x1–1 = 1

x^{0} = 1

It is often the case that a function satisfies this form but requires a bit of reformulation before proceeding to the derivative. It is the case of roots (square, cubic, etc.) representing fractional exponents.

An exponential function (of the form a^{x} with a Σ 0):

It is very easy to confuse the exponential function a^{s} with a function of the form x^{n} since both have exponents. They are, however, quite different. In an exponential function, the exponent is a variable.

An exponential function (of the form a^{x} with a Σ 0):

It is very easy to confuse the exponential function a^{s} with a function of the form x^{n} since both have exponents. They are, however, quite different. In an exponential function, the exponent is a variable.

Given the exponential function ƒ(x) = as where a Σ 0. We have

ƒ′(x) = (as)′ = a^{s} ln(a)

**Examples**

(3^{s})′ = 3^{s} ln(3)

The function e^{x}

Let the function ƒ(x) = e^{s}. Then

ƒ′(x) = (e^{s})′ = e^{s}

Here is a special case of the previous rule since the function ƒ(x) = e^{x} is an exponential function with a=e.

Therefore ƒ′(x) = (e^{x})′ = e^{x} ln(e) = e^{x} (1) = e^{x}

The logarithmic function ln x

Given the logarithmic function ƒ(x) = ln x. We have

ƒ′(x) = (ln x)′= 1/x

**Basic derivation rules**

We will generally have to confront not only the functions presented above, but also combinations of these: multiples, sums, products, quotients and composite functions. We therefore need to present the rules that allow us to derive these more complex cases.

**Constant multiples**

Let k be a real constant and ƒ(x) any given function. Then

(k ƒ(x))*′ *= k ƒ*′*(x)

In other words, we can forget the constant which will remain unchanged and only derive the function of x.

**Solution**

We need to derive the composite function u3, where u = x2 — 4. Consequently, we need to use the chain derivative.

ƒ*′*(x) = [(x^{2} – 4)^{3}]*′** *

= 3(x^{2} – 4)^{2} . (x^{2} – 4)*′** *

= 3(x^{2} – 4)^{2} . 2x

= 6x(x^{2} – 4)^{2}

(4x^{2})′ = 4(x^{2})′ = 4(2x) = 8x

(—5e^{s})′ = —5(e^{s}) = —5e^{s}

**Addition and subtraction of functions**

Let ƒ(x) and g(x) be two functions. Then

** **(ƒ(x) ± g(x))*′ *= ƒ*′*(x) ± g*′*(x)

When we derive a sum or a subtraction of two functions, the previous rule states that the functions can be individually derived without changing the operation linking them.

**Example**

(e^{s} + x^{5})′ = (e^{s})′ + (x^{5})′ = e^{s} + 5x^{4}

**Product rule**

Let ƒ(x) and g(x) be two functions. Then the derivate of the product

(ƒ(x) g(x) )′ = ƒ′(x) g(x) + ƒ(x) g′(x)

We must follow this rule religiously and not succumb to the temptation of writing

(ƒ(x) g(x))′ = ƒ′(x) g′(x); a faulty statement.

**Example**

(x^{3} e^{s})′ = (x^{3})′e^{s} + x^{3}(es)′

= 3x^{2} e^{s} + x^{3} e^{s}

**Quotient rule**

Let ƒ(x) and g(x) be two functions. Then the derivative of the quotient

f(x) ƒ′(x)g(x) — ƒ(x)g′(x)

—– = ————————

g(x) [g(x)]^{2}

Just as with the product rule, the quotient rule must religiously be respected

## Derivative of composite functions

A composite function is a function with form ƒ(g(x)).

**How do we recognize a composite function?**

A composite function is in fact a function that contains another function. If you have a function that can be broken down into many parts, where each part is in itself a function and where these parts are not linked by addition, subtraction, product or division, you usually have a composite function.

For example, the function ƒ(x) = e^{s^}^{3} is a composite function. We can write it as

ƒ(g(x)) where g(x) = x^{3}.

Unlike the function f(x) = x^{3}e^{x} which is not a composite function. It is only the product of functions.

Here are a few examples of composite functions:

**The Chain Rule**

Let ƒ and g be two functions. Then the derivative of the composite function

ƒ(g(x)) is

(ƒ(g(x)))*′ *= ƒ*′*(g(x)) g*′*(x)

or (ƒ(u))*′ *= ƒ*′*(u) u*′ , *where u = g(x)

**Chain derivatives of usual ****functions**

In concrete terms, we can express the chain rule for the most important functions as follows :

**Examples**

[e^{sSns}]′ = e^{sSns}.(xlnx)′

= e^{sSns}[(x)′lnx + x(lnx)′] (product rule)

= e^{sSns} (1. lnx + x. ^{1})s

= esSns(lnx + 1)

**Evaluation of the slope of the tangent at one point**

As we mentioned at the very beginning, the derivative function ƒ′(x) represents the slope of the tangent line at ƒ(x) at all points x. We will often have to evaluate this slope at a specific point.

To evaluate the slope of the tangent of the function ƒ(x) at the point x = 1 for example, we most certainly cannot calculate ƒ(1) and derive this value… we would then obtain a slope of 0 since ƒ(1) is a constant. Instead, we need to find the derivative ƒ′(x) at all points and then evaluate it at x = 1. We will use the notation ƒ′(a) to represent the derivative of the function ƒ evaluated at the point x = a.

**Example**

Evaluate the slope of the functionƒ(x) = x^{3}e^{s} at the point x = 0.

We are looking to calculate ƒ′(0). We must first find the derivative at all points, ƒ′(x). Yet earlier we demonstrated that ƒ′(x) = (x^{3}e^{s})′ = 3x^{2}e^{s} + x^{3}e^{s}

Evaluated at x = 0, we obtain ƒ′(0) = 3. 0^{2}e^{0} + 0^{3}e^{0} = 0. The slope of the function ƒ(x) = x^{3}e^{s} is therefore zero at x = 0. We will let you verify that this is not the case at point x = 1.

Increasing and decreasing functions

There is a direct relationship between the growth and decline of a function and the value of its derivative at one point.

• If the value of the derivative is negative at a given point, this indicates that the function is decreasing at that point.

• If the value of the derivative is positive at a given point, this indicates that the function is increasing at that point.

**Example**

• Find the derivative of the function ƒ(x) = (x^{2} — 4)^{3}.

Solution

• We need to derive the composite function u^{3}, where u = x^{2} — 4. Consequently, we need to use the chain derivative.

ƒ′(x) = [(x^{2} – 4)^{3}]′

= 3(x^{2} – 4)^{2}. (x^{2} – 4)′

= 3(x^{2} – 4)^{2}. 2x

= 6x(x^{2} – 4)2