For `y = (ln(x))/e^x` what is `dy/dx` and `(d^2y)/(dx^2) `

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The function ` y = (ln (x))/e^x` . Use the quotient rule that gives the derivative of `f(x)/g(x)` as `(f'(x)*g(x) - f(x)*g'(x))/(g(x))^2`

`dy/dx = ((1/x)*e^x - ln(x)*e^x)/e^(2x)`

=> `(1 - x*ln x)/(x*e^x)`

`(d^2y)/(dx^2) = ((-1 - ln x)*x*e^x + (x*ln x - 1)(e^x + x*e^x))/(x^2*e^(2x))`

=> `((-1 - ln x)*x + (x*ln x - 1)(1 + x))/(x^2*e^x)`

=> `(-x - x*ln x + x*ln x - 1 + x^2*ln x - x)/(x^2*e^x)`

=> `(-2x - 1 + x^2*ln x)/(x^2*e^x)`

=> `(x^2*ln x - 2x - 1)/(x^2*e^x)`

The derivative `dy/dx = (1 - x*ln x)/(x*e^x)` and `(d^2y)/(dx^2) = (x^2*ln x - 2x - 1)/(x^2*e^x)`

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