Differentiable Function

In calculus, a differentiable function is a continuous function whose derivative exists at all points on its domain. That is, the graph of a differentiable function must have a (non-vertical) tangent line at each point in its domain, be relatively "smooth" (but not necessarily mathematically smooth), and cannot contain any breaks, corners, or cusps. Differentiability lays the foundational groundwork for important theorems in calculus such as the mean value theorem.

A function $f$ is differentiable at a point $x_0$ if

1) $f$ is continuous at $x_0$ and
2) the slope of tangent at point $x_0$ is well defined.

At point $c$ on the interval $[a, b]$ of the function $f(x)$, where the function is continuous on $[a, b]$, there is a corner if

$$\lim_{x \rightarrow c^{-} } \dfrac{d}{dx} \big[ f(x) \big] \neq \lim_{x \rightarrow c^{+} } \dfrac{d}{dx} \big[ f(x) \big]$$

but both limits exist.

Note: Corners are not differentiable.

At point $c$ on the interval $[a, b]$ of the function $f(x)$, where the function is continuous on $[a, b] ,$ there is a cusp if exactly one of the following is true:

• $\displaystyle \lim_{x \rightarrow c^{-} } \dfrac{d}{dx} \left[ f(x) \right] = + \infty$ and $\displaystyle \lim_{x \rightarrow c^{+} } \dfrac{d}{dx}\left[ f(x) \right] = + \infty,$ or
• $\displaystyle \lim_{x \rightarrow c^{-} } \dfrac{d}{dx}\left[ f(x) \right] = - \infty$ and $\displaystyle \lim_{x \rightarrow c^{+} } \dfrac{d}{dx}\left[ f(x) \right] = - \infty.$

Note: Cusps are not differentiable

True or False?

If a function is everywhere continuous, then it is everywhere differentiable.

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False.

Example 1: The Weierstrass function is infinitely bumpy, so that at no point can you take a derivative. But it's everywhere connected.

Example:2 $f(x) = \left| x \right|$ is everywhere continuous but it has a corner at $x=0.$ We cannot find the derivatives at corners because the derivative is a limit and the derivative from the left of zero is not equal to the derivative of the right of zero.

A function $f(x)$ is smooth along $(a, b)$ if the first derivative, $f'(x)$, is continuous along $(a, b).$

Mean Value Theorem For Derivatives

If a function $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least a point $x=c$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.

For the function $f(x) = x^{3}$, find the value(s) of $x$ satisfying the mean value theorem on the interval $[-1, 1]$.

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We know that $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, so there do exist $x$ values that satisfy the mean value theorem.

So, we want to know where the slope of the tangent equals the slope of the secant given by $-1$ to $1$ on $f(x) = x^3$, which is $\frac{f(1) - f(-1) }{1-(-1)} = 1.$

By the power rule, $f'(x) = 3x^{2}.$ Set $f'(x) = 1$ and solve the equation $3x^{2} = 1:$

$$3x^2 - 1 = 0 \implies x= \pm \dfrac{1}{\sqrt{3}}.$$

Those two $x$ values are the values of $x$ whose slopes of the tangents equal the slope of the secant line from $-1$ to $1$ on $f(x) = x^{3}$. $_\square$

Mean Value Theorem For Integrals

Given a function $f(x)$ on the interval $[a, b],$ if $h(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists a point $c$ between $a$ and $b$ such that

$$\int_{a}^{b} f(x)\, dx = f(c) (b-a).$$

Note: This enables you to determine the mean value of $f(x)$ on the interval $[a, b]$. Think about it this way: $f(c)$ is the average height of $f(x)$, which enables you just to multiply $f(c)$ by the width of the rectangle, $b-a$, to get the area under the curve. So, by dividing both sides by $b-a$, we can write the average value of $f(x)$ like this:

$$\text{average value}{\Large|}_{f(x)} = \dfrac{1}{b-a} \displaystyle \int_{a}^{b} f(x)\, dx.$$

Given the function $f(x) = \displaystyle \int_{0}^{x} \big(t^{3} - 3t^{2} + t + 1 \big)\, dt$, find the average value of $f(x)$ from $x = -2$ to $x = 4$.

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We have

$$\begin{aligned} \text{avg}|_{f(x)} &= \frac{1}{4-(-2)} \int_{-2}^{4} \int_{0}^{x} \left( t^{3} - 3t^{2} + t + 1 \right)\, dt \\ &= \frac{1}{6} \int_{-2}^{4} \left( \frac{x^{4}}{4} - x^3 + \frac{x^2}{2} + x \right)\, dx \\ &= \frac{1}{6} \left[ \frac{x^{5}}{20} - \frac{x^4}{4} + \frac{x^3}{6} + \frac{x^2}{2} \right]_{-2}^{4} \\ &=\frac{9}{5}. \ _\square \end{aligned}$$

We can find the length of a smooth curve $f(x)$ along a certain interval $[a, b].$ Remember local linearity and tangent approximation? If we zoom in on a curve, eventually it starts to look like a line. So, is it not fair to say that if we a draw a secant line through $\big(c, f(c)\big)$ and $\big(d, f(c)\big)$ $\big($where $c$ and $d$ are within $(a, b)\big),$ as $d-c$ $($let's call it $\Delta x)$ approaches zero, the length of the secant line $($let's denote as $L)$ is quite close to the length of the portion of curve cut off by $x = c$ and $x = d?$ Additionally, if we have a $\Delta x$, then we must have a $\Delta y$, such that $\left( \Delta x \right)^2 + \left( \Delta y \right)^2 = L^2$. Then $L = \sqrt{ \left( \Delta x \right)^2 + \left( \Delta y \right)^2 }$. We can then sum all of these curve lengths $L$ together with regular partitions of $\Delta x$ and have an infinite number of curve lengths. So, we write

$$\displaystyle \lim_{n \rightarrow \infty} \sum_{k = 1}^{n} \sqrt{\left( \Delta x \right)^2 + \left( \Delta y \right)^2},$$ where $k$ is the $n^\text{th}$ partition along $(a, b),$ and there is an infinite number of partitions along $(a, b).$

In calculus, what do we like to do with sums? Convert them into integrals! But do we have a Riemann sum? No. So we can't make it an integral quite yet. To get a Riemann sum, we need to have an $f_{k}^{*}$, the height of a rectangle at any given partition, and $\Delta x$, the width of a regular partition. But, we can factor out $\left( \Delta x \right)^{2}$. So we get

$$\displaystyle \lim_{n \rightarrow \infty} \sum_{k = 1}^{n} \sqrt{\left( \Delta x \right)^2 \left(1 + \frac{\left( \Delta y \right)^2}{\left( \Delta x \right)^2 }\right) },$$

which can be rewritten as

$$\displaystyle \lim_{n \rightarrow \infty} \sum_{k = 1}^{n} \sqrt{1 + \left( \frac{ \Delta y }{\Delta x } \right)^2 } ~\Delta x.$$

We can then write $\dfrac{ \Delta y }{\Delta x }$ as $f'\left( x_{k}^{*} \right)$.

Why are we allowed to do this? The mean value theorem for derivatives states that there is some point on the interval $x=c$ on $(a, b)$ such that the slope of the secant line from $a$ to $b$ equals $f'(c)$. Except here, we replace $c$ with $x_{k}^{*}$, because that's some point on our regular partition and we want a Riemann sum.

So now, we have our Riemann sum

$$\displaystyle \lim_{n \rightarrow \infty} \sum_{k = 1}^{n} \sqrt{1 + \left[ f'\left(x_{k}^{*} \right) \right]^2 } ~\Delta x,$$

and we now have our arc length formula:

$$\boxed{\displaystyle \int_{a}^{b} \sqrt{1+\big[f'(x)\big]^2 } \, dx }.$$

$$$$ Conceptual questions to consider:

• Why does the function have to be smooth in order to use the arc length formula?
• Didn't we say that $\sqrt{x^2} = \left| x \right|?$ And absolute value returns a positive value always. So then, how do we find arc length when $\Delta x$ is negative?
• How can we derive a formula for the arc length of a certain interval on a parametric curve? What conditions are necessary for the formula to work?

What is the length of the curve $y = \sqrt[3]{x}$ on $[-8, 8]?$

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At first glance, you might say that $y$ is not smooth, because $y'$ doesn't exist at $x=0$. However, what if we make the curve smooth? Isn't it true that we can rewrite the function as $x= y^3?$ If we look at $x = y^3$, is that smooth? In a way, yes, because now $\left. \frac{dx}{dy} \right|_{y=0} = 0$, so we're good to go, and we can integrate in terms of $y$.

We need to know $\frac{dx}{dy}$, which is $3y^2$ by the power rule, and thus the formula is

$$\int_{-2}^{2} \sqrt{1+ \left[ 3y^2 \right]^{2} }\, dy.$$

Using the calculator to evaluate this integral, we can find that the length of the curve on that interval comes out to about $17.2607...$. $_\square$