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• Notes

Show that $\frac{d}{d x}\left[\tan ^{-1} x\right]=\frac{1}{x^{2}+1}$

$\tan ^{-1} x=\frac{1}{x^{2}+1}$

$y=\tan ^{-1}(x)$

$\tan y=x$

$\sec ^{2} y \cdot \frac{d y}{d x}=1$

$\frac{d y}{d x}=\frac{1}{\sec ^{2} y}$

$\sec ^{2} y=1+\tan ^{2} y$

$\frac{d y}{d x}=\frac{1}{1+\tan ^{2} y}$

$\tan ^{2} y=x^{2}$

$\rightarrow \frac{d\left(\tan ^{-1} x\right)}{d x}=\frac{1}{1+x^{2}}$

Find the derivative for the following functions

$y=\left(\tan ^{-1}(x)\right)^{2} \quad y=\tan ^{-1} x^{2} \quad f(x)=x \sin ^{-1} x+\sqrt{1-x^{2}}$

$y=\left(\tan ^{-1}(x)\right)^{2} \rightarrow y^{\prime}=2\left(\tan ^{-1}(x)\right) \cdot \frac{d}{d x}\left(\tan ^{-1}(x)\right)$

$y^{\prime}=2 \tan ^{-1}(x) \cdot \frac{1}{1+x^{2}} \rightarrow y^{\prime}=\frac{2 \tan ^{-1}(x)}{1+x^{2}}$

$y=\tan ^{-1} x^{2} \rightarrow y^{\prime}=\frac{2 x}{1+\left(x^{2}\right)^{2}}=\frac{2 x}{1+x^{4}}$

$f(x)=x \sin ^{-1} x+\sqrt{1-x^{2}} \rightarrow$

$f^{\prime}(x)=(1) \sin ^{-1} x+\frac{1}{\sqrt{1-x^{2}}}(x)+\frac{-2 x}{2 \sqrt{1-x^{2}}}$

$f^{\prime}(x)=\sin ^{-1}(x)+\frac{x}{\sqrt{1-x^{2}}}+\frac{x}{\sqrt{1-x^{2}}}$

$=\sin ^{-1}(x)+\frac{2 x}{\sqrt{1-x^{2}}}$

Find the derivative for the following functions

$h(t)=\cos ^{-1} t+\cot ^{-1}(3 / t)$

$h(t)=\cos ^{-1} t+\cot ^{-1}(3 / t)$

$h^{\prime}(t)=\frac{-1}{\sqrt{1-x^{2}}}+\frac{-\frac{-3}{t^{2}}}{1+(3 / t)^{2}}$

$h^{\prime}(t)=\frac{-1}{\sqrt{1-x^{2}}}+\frac{3 / t^{2}}{1+9 / t^{2}}$