if f(x)= 1/√{(x^2+a^2)+√(x^2+b^2)},then f'(x) is equal to

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Raj · Answer

Suppose f(x) = 1/g(x) Therefore f'(x) = -1/ x^2 Then diff.g(x) w.r.t to x Then solve the equation And at last we got the answer

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if f(x)= 1/√{(x^2+a^2)+√(x^2+b^2)},then f'(x) is equal to

if f(x)= 1/√{(x^2+a^2)+√(x^2+b^2)},then f'(x) is equal to

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if f(x)= 1/√{(x^2+a^2)+√(x^2+b^2)},then f'(x) is equal to

if f(x)= 1/√{(x^2+a^2)+√(x^2+b^2)},then f'(x) is equal to

1 Answers

Think You Can Provide A Better Answer ?

Other Related Questions on Differential Calculus

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