## Differentiation --- ## Difference quotient #### Definition Let $I$ be an open interval, $x,a\in I$, with $x\neq a$, and $f:I\to \mathbb R$. Then $\displaystyle \frac{f(x)-f(a)}{x-a}$ is the *difference quotient* of $f$ at $x$ and $a$. [Excel file demonstrating the difference quotient](https://nc.dasmithmaths.com/index.php/s/WprS2HSww9MdHaP) --- ## Derivative #### Definition Let $I$ be an open interval, $a\in I$, and $f:I\to \mathbb R$. Then $\displaystyle \lim\_{x\to a} \frac{f(x)-f(a)}{x-a}$ is the *derivative* of $f$ at $a$. So the derivative at a point is the limit of the difference quotient as $x\to a$. It is the limit of the slope of the chord between $x$ and $a$ as $x\to a$. It is the slope of the tangent at $a$. --- ## Derivative #### Definition Let $I$ be an open interval, $a\in I$, and $f:I\to \mathbb R$. Then $\displaystyle \lim\_{x\to a} \frac{f(x)-f(a)}{x-a}$ is the *derivative* of $f$ at $a$. So the derivative at a point is the limit of the difference quotient as $x\to a$. It is the limit of the slope of the chord between $x$ and $a$ as $x\to a$. It is the slope of the tangent at $a$. Note $\displaystyle \lim\_{x\to a} \frac{f(x)-f(a)}{x-a} = \lim\_{h\to 0} \frac{f(a+h)-f(a)}{h}$, using $h=x-a$. --- ## Derivative #### Definition Let $I$ be an open interval, $a\in I$, and $f:I\to \mathbb R$. Then $\displaystyle \lim\_{x\to a} \frac{f(x)-f(a)}{x-a}$ is the *derivative* of $f$ at $a$. Note $\displaystyle \lim\_{x\to a} \frac{f(x)-f(a)}{x-a} = \lim\_{h\to 0} \frac{f(a+h)-f(a)}{h}$, using $h=x-a$. #### Example The derivative of $f(x)=x^3+2$ at $a=4$ is $48$. #### Proof $\displaystyle \lim\_{h\to 0} \frac{f(a+h)-f(a)}{h} = \lim\_{h\to 0} \frac{(4+h)^3+2-4^3-2}{h} = \lim\_{h\to 0} \frac{3\times4^2h+3\times4h^2+h^3}{h} = \lim\_{h\to 0} \left(3\times4^2+3\times4h+h^2\right) = 48$. --- ## Differentiability #### Definition Let $I$ be an open interval, $a\in I$, and $f:I\to \mathbb R$. Then $\displaystyle \lim\_{x\to a} \frac{f(x)-f(a)}{x-a}$ is the *derivative* of $f$ at $a$. #### Definition Let $I$ be an open interval, $a\in I$, and $f:I\to \mathbb R$. If the derivative of $f$ at $a$ exists, then we say that $f$ is *differentiable* at $a$. If, for all $x\in I$, the derivative of $f$ at $x$ exists, then we say that $f$ is *differentiable* on the interval $I$. --- ## Differentiability #### Earlier example $\displaystyle \lim\_{h\to 0} \frac{f(a+h)-f(a)}{h} = \lim\_{h\to 0} \frac{(4+h)^3+2-4^3-2}{h} = \lim\_{h\to 0} \frac{3\times4^2h+3\times4h^2+h^3}{h} = \lim\_{h\to 0} \left(3\times4^2+3\times4h+h^2\right) = 48$, so the derivative of $f(x)=x^3+2$ at $4$ is $48$. #### Definition Let $I$ be an open interval, $a\in I$, and $f:I\to \mathbb R$. If the derivative of $f$ at $a$ exists, then we say that $f$ is *differentiable* at $a$. If, for all $x\in I$, the derivative of $f$ at $x$ exists, then we say that $f$ is *differentiable* on the interval $I$. #### Example $f(x)=x^3+2$ is differentiable at $4$. --- ## Differentiability #### Earlier example $\displaystyle \lim\_{h\to 0} \frac{f(a+h)-f(a)}{h} = \lim\_{h\to 0} \frac{(4+h)^3+2-4^3-2}{h} = \lim\_{h\to 0} \frac{3\times4^2h+3\times4h^2+h^3}{h} = \lim\_{h\to 0} \left(3\times4^2+3\times4h+h^2\right) = 48$, so the derivative of $f(x)=x^3+2$ at $4$ is $48$. #### Definition Let $I$ be an open interval, $a\in I$, and $f:I\to \mathbb R$. If the derivative of $f$ at $a$ exists, then we say that $f$ is *differentiable* at $a$. If, for all $x\in I$, the derivative of $f$ at $x$ exists, then we say that $f$ is *differentiable* on the interval $I$. #### Example $f(x)=x^3+2$ is differentiable on $\mathbb R$. Try leaving $a$ general in the above argument. --- ## Differentiability #### Earlier example $\displaystyle \lim\_{h\to 0} \frac{f(a+h)-f(a)}{h} = \lim\_{h\to 0} \frac{(4+h)^3+2-4^3-2}{h} = \lim\_{h\to 0} \frac{3\times4^2h+3\times4h^2+h^3}{h} = \lim\_{h\to 0} \left(3\times4^2+3\times4h+h^2\right) = 48$, so the derivative of $f(x)=x^3+2$ at $4$ is $48$. #### Earlier example at general point $\displaystyle \lim\_{h\to 0} \frac{f(a+h)-f(a)}{h} = \lim\_{h\to 0} \frac{(a+h)^3+2-a^3-2}{h} = \lim\_{h\to 0} \frac{3\times a^2h+3\times ah^2+h^3}{h} = \lim\_{h\to 0} \left(3\times a^2+3\times ah+h^2\right) = 3a^2$, so the derivative of $f(x)=x^3+2$ at $a$ is $3a^2$. So, for any $a\in\mathbb R$, the function $f(x)=x^3+2$ is differentiable at $a$ with derivative $3a^2$. --- ## Derivative function #### Earlier example $\displaystyle \lim\_{h\to 0} \frac{f(a+h)-f(a)}{h} = \lim\_{h\to 0} \frac{(4+h)^3+2-4^3-2}{h} = \lim\_{h\to 0} \frac{3\times4^2h+3\times4h^2+h^3}{h} = \lim\_{h\to 0} \left(3\times4^2+3\times4h+h^2\right) = 48$, so the derivative of $f(x)=x^3+2$ at $4$ is $48$. #### Earlier example at general point $\displaystyle \lim\_{h\to 0} \frac{f(a+h)-f(a)}{h} = \lim\_{h\to 0} \frac{(a+h)^3+2-a^3-2}{h} = \lim\_{h\to 0} \frac{3\times a^2h+3\times ah^2+h^3}{h} = \lim\_{h\to 0} \left(3\times a^2+3\times ah+h^2\right) = 3a^2$, so the derivative of $f(x)=x^3+2$ at $a$ is $3a^2$. So, for any $a\in\mathbb R$, the function $f(x)=x^3+2$ is differentiable at $a$ with derivative $3a^2$. --- ## Derivative function #### Earlier example at general point $\displaystyle \lim\_{h\to 0} \frac{f(a+h)-f(a)}{h} = \lim\_{h\to 0} \frac{(a+h)^3+2-a^3-2}{h} = \lim\_{h\to 0} \frac{3\times a^2h+3\times ah^2+h^3}{h} = \lim\_{h\to 0} \left(3\times a^2+3\times ah+h^2\right) = 3a^2$, so the derivative of $f(x)=x^3+2$ at $a$ is $3a^2$. So, for any $a\in\mathbb R$, the function $f(x)=x^3+2$ is differentiable at $a$ with derivative $3a^2$. We say *the derivative function* of $f$ is $f^\prime(a) = 3a^2$ or, in more common notation, $f'(x) = 3x^2$. Because $f$ is differentiable on $\mathbb R$, the domain of $f'$ is $\mathbb R$. --- ## Differentiability and continuity #### Theorem If $f$ is differentiable at a point, then $f$ is continuous at that point. So all differentiable functions are continuous functions. --- ## Differentiability and continuity #### Theorem If $f$ is differentiable at a point, then $f$ is continuous at that point. So all differentiable functions are continuous functions. Is the converse true? In other words, are all continuous functions differentiable? --- ## Differentiability and continuity #### Theorem If $f$ is differentiable at a point, then $f$ is continuous at that point. So all differentiable functions are continuous functions. Is the converse true? In other words, are all continuous functions differentiable? We know the absolute value function is continuous. But is it differentiable? [Excel file demonstrating the difference quotient](https://nc.dasmithmaths.com/index.php/s/WprS2HSww9MdHaP) --- ## Differentiability and continuity #### Theorem If $f$ is differentiable at a point, then $f$ is continuous at that point. So all differentiable functions are continuous functions. Is the converse true? In other words, are all continuous functions differentiable? We know the absolute value function is continuous. But is it differentiable? [Excel file demonstrating the difference quotient](https://nc.dasmithmaths.com/index.php/s/WprS2HSww9MdHaP) So a function can be continuous on an interval but not differentiable at a point on that interval. See the Weierstrass function for a stronger example, which is continuous on $\mathbb R$, but differentiable nowhere. --- ## Differentiation Differentiation means *finding the derivative* of a function. #### Example When we differentiate $f(x)=x^3+2$ we get $f'(x) = 3x^2$. --- ## Differentiation Differentiation means *finding the derivative* of a function. #### Example When we differentiate $f(x)=x^3+2$ we get $f'(x) = 3x^2$. There are various differentiation rules which make this easier, avoiding going back to the difference quotient for all such calculations.