# Definition of Derivative

## Related calculator: Online Derivative Calculator with Steps

There is one limit that is used very frequently in applications of calculus and different sciences. This limit has form `lim_(h->0)(f(x+h)-f(x))/h` and has special notation.

**Definition.** The **derivative** of function `f` is a function that is denoted by `f'(x)` and calculated as `f'(x)=lim_(h->0)(f(x+h)-f(x))/h`.

Process of finding derivative is called **differentiating.**

There are different notations for derivative.

Leibniz notation: `(dy)/(dx)`, `(df)/(dx)`, `d/(dx)f(x)`. Note, that here `(dy)/(dx)` is not ratio - it is just synonym for `f'(x).` It is very useful and suggestive notation, derivative is defined as a limit of ratio. If we want to indicate the value of a derivative in Leibniz notation at a specific number `a`, we use the notation `(dy)/(dx)|_(x=a)`.

Lagrange notation: `y'`, `f'(x)`.

Cauchy notation: `Dy`, `Df(x)`.

In Lagrange and Cauchy notations if we need to explicitly state with respect to what variable we want to take derivative, we write `y'_x`, `f'_x`, `D_xy`, `D_x f`.

We will mainly use Lagrange notation.

**Example 1.** If `f(x)=x^2`, find `f'(x)` and `f(3)`.

We have that `f(x+h)=(x+h)^2`.

Thus,

`f'(x)=lim_(h->0)(f(x+h)-f(x))/h=lim_(h->0)((x+h)^2-(x)^2)/(h)=lim_(h->0)(x^2+2xh+h^2-x^2)/h=`

`=lim_(h->0)(2xh+h^2)/h=lim_(h->0)(2x+h)=2x`.

So, `f'(x)=2x` and `f'(3)=2*3=6`.

**Example 2.** Find `f'(x)` if `f(x)=sqrt(x)` and state domain of `f'`.

`f'(x)=lim_(h->0)(sqrt(x+h)-sqrt(x))/h=lim_(h->0)(sqrt(x+h)-sqrt(x))/h (sqrt(x+h)+sqrt(x))/(sqrt(x+h)+sqrt(x))=lim_(h->0)(x+h-x)/(h(sqrt(x+h)+sqrt(x)))=`

`=lim_(h->0)1/(sqrt(x+h)+sqrt(x))=1/(sqrt(x+0)+sqrt(x))=1/(2sqrt(x))`

Thus, domain of `f'` is `(0,oo)`. As can be seen it is smaller then domain of `f(x)=sqrt(x)`.

What if function is given by table of values?

**Example 3.** Estimate `f'(1)` and `f'(1.5)` if

`x` | 1 | 1.5 | 2 |

`f(x)` | 2 | 3 | 7 |

We have that `f'(x)=lim_(h->0)(f(x+h)-f(x))/h`. But if we take `h=0.5` we will obtain following approximation: `f'(x)~~(f(x+0.5)-f(x))/0.5=2(f(x+0.5)-f(x))`.

Thus,

`f'(1)~~2(f(1+0.5)-f(1))=2(f(1.5)-f(1))=2(3-2)=2`.

`f'(1.5)~~2(f(1.5+0.5)-f(1.5))=2(f(2)-f(1.5))=2(7-3)=8`.

**Definition.** Function `f` is differentiable at `a` if `f'(a)` exists.

**Definition.** Function `f` is differentiable on open interval `(a,b)` or `(a,oo)` or `(-oo,a)` or `(-oo,oo)` if it is differentiable at every number in that interval.

**Example 4.** Where is `f(x)=|x|` is differentiable?

If `x>0`, then `|x|=x` and we can choose `h` small enough so that `x+h>0` and hence `|x+h|=x+h`.

Therefore, for `x>0` `f'(x)=lim_(h->0)(|x+h|-|x|)/h=lim_(h->0)(x+h-x)/h=lim_(h->0)1=1` and so `f` is differentiable for any `x>0`.

Similarly, if `x<0`, then `|x|=-x` and we can choose `h` small enough so that `x+h<0` and hence `|x+h|=-(x+h)`.

Therefore, for `x<0` `f'(x)=lim_(h->0)(|x+h|-|x|)/h=lim_(h->0)(-(x+h)-(-x))/h=lim_(h->0)-1=-1` and so `f` is differentiable for any `x<0`.

For `x=0` we have to investigate`f'(0)=lim_(h->0)(|0+h|-|0|)/h=lim_(h->0)(|h|)/h`.

Let's compute one-sided limits:

`lim_(h->0^+)|h|/h=lim_(h->0^+)h/h=lim_(h->0^+)1=1`

`lim_(h->0^-)|h|/h=lim_(h->0^-)-h/h=lim_(h->0^-)-1=-1`

Since `lim_(h->0^+)(|h|)/h!=lim_(h->0^-)(|h|)/h` then`lim_(h->0)(|h|)/h` doesn't exist and thus `f(x)` is not differentiable at `x=0.`

Therefore, `|x|` is differentiable at all `x` except `0`.

**Theorem.** If `f` is differentiable at `a` then `f` is continuous at `a`. Converse is not true.

Proof. Since `f` is differentiable at `a` then `f'(a)=lim_(h->0)(f(a+h)-f(a))/h`.

If we make substitution `x=a+h` then `h=x-a`. As `h->0` `x->a`.

Therefore, `f'(a)=lim_(x->a)(f(x)-f(a))/(x-a)`.

Now consider identity `f(x)-f(a)=f(x)-f(a)` or equivalently `f(x)-f(a)=(f(x)-f(a))/(x-a)(x-a)`. Taking limits of both sides gives `lim_(x->a)(f(x)-f(a))=lim_(x->a)((f(x)-f(a))/(x-a)(x-a))`.

Using property of limit of product we obtain that`lim_(x->a)(f(x)-f(a))=lim_(x->a)(f(x)-f(a))/(x-a)lim_(x->a)(x-a)` or `lim_(x->a)(f(x)-f(a))=f'(a)*(a-a)`.

Thus, `lim_(x->a)(f(x)-f(a))=0`.

Now, `lim_(x->a)f(x)=lim_(x->a)(f(x)-f(a)+f(a))=`

`=lim_(x->a)(f(x)-f(a))+lim_(x->a)(f(a))=0+f(a)=f(a)`.

Therefore `lim_(x->a)f(x)=f(a)` which means that function is continuous at `a`.

Converse is not true: if `f` is continuous at `a` then it doesn't mean that `f` is differentiable at `a`. Consider function `f(x)=|x|`. It is continuous at `x=0`, but as was shown in example 4 it is not differentiable at 0.