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EXERCISE 1.4 · Q169

Q.If x=log⁡(1+t2)x=\log(1+t^2), y=t−tan⁡−1ty=t-\tan^{-1}t, show that dydx=ex−12\dfrac{dy}{dx}=\dfrac{\sqrt{e^x-1}}{2}.

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We have x=log⁡(1+t2)x=\log(1+t^2), y=t−tan⁡−1ty=t-\tan^{-1}t.

Step 1.

dxdt=2t1+t2\frac{dx}{dt}=\frac{2t}{1+t^2}

Step 2.

dydt=1−11+t2=(1+t2)−11+t2=t21+t2\frac{dy}{dt}=1-\frac{1}{1+t^2}=\frac{(1+t^2)-1}{1+t^2}=\frac{t^2}{1+t^2}

Step 3.

dydx=t21+t22t1+t2=t22t=t2\frac{dy}{dx}=\frac{\dfrac{t^2}{1+t^2}}{\dfrac{2t}{1+t^2}}=\frac{t^2}{2t}=\frac{t}{2} …

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