InterviewSolution
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InterviewSolution
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Find `(dy)/(dx)`for the function:`y=a^(sin^(-1)x)^2` |
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Answer» Correct Answer - `(2log a sin^(-1)x)/(sqrt(1-x^(2)))a(sin^(-1)x)^(2)` `"Let "y=a^((sin^(-1)x)^(2))`. Using chain rule, we get `(dy)/(dx)=(d)/(dx){a^((sin^(-1)x)^(2))}` `=a^((sin^(-1)x)^(2))loga(d)/(x){(sin^(-1)x)^(2)}` `=a^((sin^(-1)x)^(2))(log a) 2 (sin ^(-1)x)^(1)(d)/(dx)(sin^(-1)x)` `=a^((sin^(-1)x)^(2))(log a)2 sin^(-1)x(1)/(sqrt(1-x^(2)))` `=(2 log a sin^(-1)x)/(sqrt(1-x^(2)))a^((sin^(-1)x)^(2))` |
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