Implicit differentiation (HL)

Using this property over and over

\frac{\text dx}{dy} = \left(\frac{\text dy}{\text dx}\right)^{-1}

Implicit differentiation depends on the point $(x, y)$ , as opposed to regular differentiation only depending on $x$ .

Rules

f(x, y) = g(x, y)^\ddagger \implies \frac{\text d}{\text dx}f(x, y) = \frac{\text d}{\text dx}g(x, y)

$^\ddagger$ Caution should be taken so that you don’t take $x = 1$ and output $1 = 0$ . Typically the presence of a $y$ indicates it is safe to implicitly differentiate. In other words, implicit differentiation does not work on ${f(x) = g(x)}$ .

\frac{\text d}{\text dx}f(y) = \frac{\text df}{\text dy}\cdot\frac{\text dy}{\text dx}

\frac{\text dx}{dy} = \left(\frac{\text dy}{\text dx}\right)^{-1}

Derivation: Differentiate $\arctan x$ with respect to $x$ .

$\arctan$ has a range $y\in\left]-\frac{\pi}{2}, \frac{\pi}{2}\right[$ , which is the domain of the corresponding branch of $\tan$ .

\begin{align*} \tan y &= x, \;\; y\in\left]-\frac{\pi}{2}, \frac{\pi}{2}\right[ \\ \frac{\d}{\d x} \tan y &= \frac{\d}{\d x}x \\ \frac{\text dy}{\text dx} \cdot \sec^2 y &= 1 \\ \frac{\text dy}{\text dx} &= \cos^2 y \end{align*}

opposite is $x$ , adjacent is $1$ , hypotenuse (using Pythagorean theorem) is $\pm\sqrt{1 + x^2}$

\begin{align*}\frac{\text dy}{\text dx} &= \left(\pm\frac{1}{\sqrt{1+x^2}}\right)^2 \\ \frac{\text d}{\text dx} \arctan x &= \frac{1}{1+x^2}\qed\end{align*}