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Question
If y=cos−1(xn−x−nxn+x−n) then y′(x) is equal to
If y=cos−1(xn−x−nxn+x−n) then y′(x) is equal to
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Solution
The correct option is D −2nxn|x|(x2n+1) if n is odd
Given y=cos−1(xn−x−nxn+x−n)Differentiating both sides w.r.t x we gety′(x)=−1√1−(x2n−1x2n+1)2ddx(x2n−1x2n+1)
=x2n+1√(x2n+1)2−(x2n−1)2×2nx2n−1(x2n+1)−2nx2n−1(x2n−1)(x2n+1)2
=−1√4x2n.4nx2n−1(x2n+1)=−2nx2n−1|xn|(x2n+1)
=−2n(|x|2)nx|x|n(x2n+1)=−2n|x|nx(x2n+1)
=−2nxn−1x2n+1 if n is even and =−2nxn|x|(x2n+1) if n is odd
Given y=cos−1(xn−x−nxn+x−n)
Differentiating both sides w.r.t x we get
y′(x)=−1√1−(x2n−1x2n+1)2ddx(x2n−1x2n+1)
=x2n+1√(x2n+1)2−(x2n−1)2×2nx2n−1(x2n+1)−2nx2n−1(x2n−1)(x2n+1)2
=−1√4x2n.4nx2n−1(x2n+1)=−2nx2n−1|xn|(x2n+1)
=−2n(|x|2)nx|x|n(x2n+1)=−2n|x|nx(x2n+1)
=−2nxn−1x2n+1 if n is even and =−2nxn|x|(x2n+1) if n is odd
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