Differentiate √x2+1+√x2−1√x2+1−√x2−1 with respect to x .

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Let y=√x2+1+√x2−1√x2+1−√x2−1 ⇒y=√x2+1+√x2−1√x2+1−√x2−1×√x2+1+√x2−1√x2+1+√x2−1 ⇒y=(√x2+1+√x2−1)2(√x2+1−√x2−1)(√x2+1+√x2−1) ⇒y=(√x2+1)2+(√x2−1)2+2√(x2+1)(x2−1)(√x2+1)2−(√x2−1)2 ⇒y=x2+1+x2−1+2√(x2)2−(1)2(x2+1)−(x2−1) ⇒y=2x2+2√x4−12 ⇒y=x2+√x4−1 On differentiating y with respect to x, we get dydx=ddx(x2+√x4−1) ⇒dydx=ddx(x2)+ddx(√x4−1) ⇒dydx=ddx(x2)+ddx[(x4−1)12] We know ddx(xn)=nxn−1 ⇒dydx=2x+12(x4−1)12−1ddx(x4−1) ⇒dydx=2x+12(x4−1)12ddx(x4−1) ⇒dydx=2x+12√x4−1[ddx(x4)−ddx(1)] We have ddx(x4)=4x3 and derivative of a constant is 0 . ⇒dydx=2x+12√x4−1[4x3−0] ⇒dydx=2x+12√x4−1×4x3 therefore dydx=2x+2x3√x4−1 Thus, ddx(√x2+1+√x2−1√x2+1−√x2−1)=2x+2x3√x4−1

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