Find the derivative of y = log x + 1/x with respect to x.

Question

TextDerivative of Sum and Difference of FunctionsDerivative of Logarithmic Functions

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If y = log √(sec √x), then the value of dy/dx at x = π²/16 is :

If y = log (sin eˣ), then dy/dx is :

If y = log[tan(π/4 + x/2)], then dy/dx is :

MCQ (Single)Derivative of Logarithmic Functions

(a) If y√(x² + 1) = log(√(x² + 1) – x), show that (x² + 1) dy/dx + xy + 1 = 0.

If y√(x² + 1) = log(√(x² + 1) − x), show that (x² + 1) dy/dx + xy + 1 = 0.

If x = e^(y/x), then prove that dy/dx = (x-y)/(x log x).