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Exercise 10.5 · Q8

Q.If f(x)=xtan⁡−1xf(x) = x\tan^{-1}x, then f′(1)f'(1) is

(1) 1+π41+\dfrac{\pi}{4}
(2) 12+π4\dfrac12+\dfrac{\pi}{4}
(3) 12−π4\dfrac12-\dfrac{\pi}{4}
(4) 22
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Step 1. f(x)=xtan⁡−1xf(x)=x\tan^{-1}x. Apply the product rule with u=xu=x, v=tan⁡−1xv=\tan^{-1}x:

f′(x)=tan⁡−1x+x⋅11+x2f'(x)=\tan^{-1}x+x\cdot\frac{1}{1+x^2}

Step 2. Evaluate at x=1x=1:

f′(1)=tan⁡−1(1)+1⋅11+12=π4+12f'(1)=\tan^{-1}(1)+1\cdot\frac{1}{1+1^2}=\frac{\pi}{4}+\frac12 …

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