1
Question
Differentiate : tan−1(√x2+a2+x√x2+a2−x) w.r.t x
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Solution
y=tan−1(√x2+a2+x√x2+a2−x)dydx=11+(√x2+a2+x√x2+a2−x)2ddx(√x2+a2+x√x2+a2−x)=(√x2+a2−x)2(√x2+a2−x)2+(√x2+a2+x)2×(√x2+a2−x)ddx(√x2+a2+x)−(√x2+a2+x)ddx(√x2+a2−x)(√x2+a2−x)2=(√x2+a2−x)2(√x2+a2−x)2+(√x2+a2+x)2×(√x2+a2−x)(2x2√x2+a2+1)−(√x2+a2+x)(2x2√x2+a2−1)(√x2+a2−x)2=(√x2+a2−x)2(√x2+a2−x)2+(√x2+a2+x)2×(√x2+a2−x)(x√x2+a2+1)−(√x2+a2+x)(x√x2+a2−1)(√x2+a2−x)2
=x√x2+a2(√x2+a2−x−√x2+a2−x)+1(√x2+a2−x+√x2+a2+x)(√x2+a2+x)2+(√x2+a2−x)2
=x√x2+a2×−2x+2√x2+a2(√x2+a2+x)2+(√x2+a2−x)2
=−2x2+2x√x2+a2(√x2+a2)(x2+a2+x2−2x√x2+a2+x2+a2+x2+2x√x2+a2)=−2x2+2x√x2+a2(√x2+a2)(4x2+2a2)
=−x2+x√x2+a2(√x2+a2)(2x2+a2)
y=tan−1(√x2+a2+x√x2+a2−x)
dydx=11+(√x2+a2+x√x2+a2−x)2ddx(√x2+a2+x√x2+a2−x)
=(√x2+a2−x)2(√x2+a2−x)2+(√x2+a2+x)2×(√x2+a2−x)ddx(√x2+a2+x)−(√x2+a2+x)ddx(√x2+a2−x)(√x2+a2−x)2
=(√x2+a2−x)2(√x2+a2−x)2+(√x2+a2+x)2×(√x2+a2−x)(2x2√x2+a2+1)−(√x2+a2+x)(2x2√x2+a2−1)(√x2+a2−x)2
=(√x2+a2−x)2(√x2+a2−x)2+(√x2+a2+x)2×(√x2+a2−x)(x√x2+a2+1)−(√x2+a2+x)(x√x2+a2−1)(√x2+a2−x)2
=x√x2+a2(√x2+a2−x−√x2+a2−x)+1(√x2+a2−x+√x2+a2+x)(√x2+a2+x)2+(√x2+a2−x)2
=x√x2+a2×−2x+2√x2+a2(√x2+a2+x)2+(√x2+a2−x)2
=−2x2+2x√x2+a2(√x2+a2)(x2+a2+x2−2x√x2+a2+x2+a2+x2+2x√x2+a2)
=−2x2+2x√x2+a2(√x2+a2)(4x2+2a2)
=−x2+x√x2+a2(√x2+a2)(2x2+a2)
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