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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

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

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

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

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

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