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If, y = tan^-1(x/a), then what is the value of y’?(a) (2ax/(x^2 – a^2)^2)(b) -(2ax/(x^2 – a^2)^2)(c) (2ax/(x^2 + a^2)^2)(d) -(2ax/(x^2 + a^2)^2)I got this question in unit test.Origin of the question is Second Order Derivative topic in division Limits and Derivatives of Mathematics – Class 11 |
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Answer» CORRECT CHOICE is (d) -(2ax/(x^2 + a^2)^2) Easy explanation: We have, y = tan^-1(x/a) Differentiating two times, with respect to x, we get, y’ = d/dx(tan^-1(x/a)) = 1/(1 + (x/a)^2)*d/dx(x/a) = a^2/(x^2 + a^2)* 1/a = a/(x^2 + a^2) y” = d/dx (a/(x^2 + a^2)) = a* d/dx(x^2 + a^2)^-1 = a(-1) (x^2 + a^2)^-2*d/dx(x^2 + a^2) Now solving we get, = -a/(x^2 + a^2)^2*2X = -(2ax/(x^2 + a^2)^2) |
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