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EXERCISE 1.2 · Q102

Q.Differentiate the following w.r.t. xx: tan⁡−12x1+22x+1\tan^{-1}\dfrac{2^x}{1+2^{2x+1}}

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Let s=2xs=2^x, so 22x+1=2s22^{2x+1}=2s^2, and y=tan⁡−1(s1+2s2)y=\tan^{-1}\left(\dfrac{s}{1+2s^2}\right). With N=s,D=1+2s2N=s,D=1+2s^2 (functions of xx via s=2xs=2^x): N′=sln⁡2N'=s\ln2, and since D=1+2s2D=1+2s^2, D′=4s⋅sln⁡2=4s2ln⁡2D'=4s\cdot s\ln2=4s^2\ln2. So N′D−ND′=sln⁡2(1+2s2)−s⋅4s2ln⁡2=sln⁡2[(1+2s2)−4s2]=sln⁡2(1−2s2)N'D-ND'=s\ln2(1+2s^2)-s\cdot4s^2\ln2=s\ln2[(1+2s^2)-4s^2]=s\ln2(1-2s^2). Also D2+N2=(1+2s2)2+s2=1+5s2+4s4=(1+4s2)(1+s2)D^2+N^2=(1+2s^2)^2+s^2=1+5s^2+4s^4=(1+4s^2)(1+s^2). So $\dfrac{dy}{dx}=\dfr …

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