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“Differentiability in Depth: Piecewise Functions and Functional Equations” is available to Standard subscribers.

Beyond the textbook. Find where absolute-value, greatest-integer and piecewise functions fail to be differentiable. See what a functional equation such as f(x + y) = f(x) + f(y) forces when f is continuous at one point. Then find patterns in the n-th derivative with Leibniz's rule. It is at the level of JEE Advanced, A-Level Further Maths and IB Higher Level, with a full timed paper.

3Lessons12Problems2Python Labs1Question Set1Mock Test

Read the first lesson freeWhere a function is not differentiable

What's inside

Where a function is not differentiable: A function is not differentiable at a corner, a vertical tangent or a break. For $|g(x)|$ the corners sit at the simple zeros of $g$ , and a double zero leaves no corner. A product $g(x)|x - a|$ is differentiable at $a$

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Functional equations and what continuity forces: A functional equation such as

f(x + y) = f(x) + f(y)

is an equation for a whole function. Putting special values into it gives

f(0) = 0

f(-x) = -f(x)

and

f(nx) = nf(x)

. If

f

is also continuous at a single point, then it is continuous everywhere and

f(x) = cx

. Similar results hold for

f(x + y) = f(x)f(y)

and

f(xy) = f(x) + f(y)

The n-th derivative and Leibniz's rule: Some functions have an

n

-th derivative that follows a pattern. For

e^{ax}

it is

a^ne^{ax}

, for

\sin(ax + b)

it is

a^n\sin\left(ax + b + \dfrac{n\pi}{2}\right)

, and for

\dfrac{1}{x + 1}

it is

\dfrac{(-1)^nn!}{(x + 1)^{n+1}}

. For a product, Leibniz's rule gives

(uv)^{(n)} = \displaystyle\sum_{k=0}^{n}{}^nC_k\,u^{(k)}v^{(n-k)}

, with the same coefficients as the binomial theorem.