Implicit differentiation (HL)

Using this property over and over

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

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

Rules

f(x,y)=g(x,y)‡  ⟹  ddxf(x,y)=ddxg(x,y)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=1x = 1 and output 1=01 = 0. Typically the presence of a yy indicates it is safe to implicitly differentiate. In other words, implicit differentiation does not work on f(x)=g(x){f(x) = g(x)}.

ddxf(y)=dfdy⋅dydx\frac{\text d}{\text dx}f(y) = \frac{\text df}{\text dy}\cdot\frac{\text dy}{\text dx}
dxdy=(dydx)−1\frac{\text dx}{dy} = \left(\frac{\text dy}{\text dx}\right)^{-1}

Derivation: Differentiate arctan⁡x\arctan x with respect to xx.

arctan⁡\arctan has a range y∈]−π2,π2[y\in\left]-\frac{\pi}{2}, \frac{\pi}{2}\right[, which is the domain of the corresponding branch of tan⁡\tan.

tan⁡y=x,    y∈]−π2,π2[ d dxtan⁡y= d dxxdydx⋅sec⁡2y=1dydx=cos⁡2y\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 xx, adjacent is 11, hypotenuse (using Pythagorean theorem) is ±1+x2\pm\sqrt{1 + x^2}

dydx=(±11+x2)2ddxarctan⁡x=11+x2■\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*}
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