NSW Y12 Maths - Extension 1 Calculus Derivative of Inverse Functions

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Derivative of Inverse Functions Theory

$The following rules may be applied\\ $\begin{aligned} & \dfrac{d}{d x}\left(\sin ^{-1} x\right)=\dfrac{1}{\sqrt{1-x^2}},-1<x<1 \\ & \dfrac{d}{d x}\left(\sin ^{-1} \dfrac{x}{a}\right)=\dfrac{1}{\sqrt{a^2-x^2}},-a<x<a \\ & \dfrac{d}{d x}\left(\cos ^{-1} x\right)=-\dfrac{1}{\sqrt{1-x^2}},-1<x<1 \\ & \dfrac{d}{d x}\left(\cos ^{-1} \dfrac{x}{a}\right)=-\dfrac{1}{\sqrt{a^2-x^2}},-a<x<a \\ & \dfrac{d}{d x}\left(\tan ^{-1} x\right)=\dfrac{1}{1+x^2}, \text{ all } x \\ & \dfrac{d}{d x}\left(\tan ^{-1} \dfrac{x}{a}\right)=\dfrac{a}{a^2+x^2}, \text{ all } x \end{aligned}$\\ \begin{multicols}{2} \textbf{Example 1}\\ %Question 12026 Differentiate ${\rm{y = ta}}{{\rm{n}}^{ - 1}}\left( {\dfrac{1}{x}} \right)$\\ \textbf{Example 1 solution}\\ $\begin{aligned} \text {Let }u =\frac{1}{x} \rightarrow u &=x^{-1} \\ \frac{d u}{d x} &=-x^{-2} =-\frac{1}{x^{2}}\\ \end{aligned}$\\ $\begin{aligned} y&=\tan ^{-1} u \\ \frac{dy}{dx}&=\frac{1}{1+u^{2}}\\ \frac{dy}{dx} &=\frac{d y}{d u} \times \frac{d u}{dx} \\ &=\frac{1}{1+u^{2}} \times -\frac{1}{x^{2}} \\ &=\frac{1}{1+\frac{1}{x^{2}}} \times-\frac{1}{x^{2}} \\ &=\frac{x^{2}}{1+x^{2}} \times-\frac{1}{x^{2}} \\ &=\frac{-1}{1+x^{2}} \end{aligned}$\\ \columnbreak \textbf{Example 2}\\ %Question 12027 Find the gradient of the tangent to $y = {\sin ^{ - 1}}\dfrac{{3x}}{2}$ at $x = \dfrac{1}{3}$\\ \textbf{Example 2 solution}\\ $\begin{aligned} y &=\sin ^{-1} \frac{3 x}{2} \\ \text {Let } u &=\frac{3 x}{2} \quad y=\sin ^{-1} u \\ \frac{du}{dx} &=\frac{3}{2} \quad \frac{d y}{du}=\frac{1}{\sqrt{1-u^{2}}}\\ \frac{d y}{d x} &=\frac{dy}{du} \times \frac{du}{dx} \\ &=\frac{1}{\sqrt{1-u^{2}}} \times \frac{3}{2} \\ &=\frac{1}{\sqrt{1-\frac{9 x^{2}}{4}}} \times \frac{3}{2} \\ &=\frac{2}{\sqrt{4-9x^{2}}} \times \frac{3}{2}=\frac{3}{\sqrt{4-9x^{2}}} \end{aligned}$\\ ${At } x=\dfrac{1}{3}, \dfrac{dy}{dx}=\dfrac{3}{\sqrt{4-1}}=\dfrac{3}{\sqrt{3}}=\sqrt{3}$\\ \end{multicols}$

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