Tangent of a function

Calculus Level pending

Find the gradient of the tangent of the function $e^{x} \ln { y } = 1$ at the point $(0,e).$

2 e

- e

-2 e

e

1 solution

Karan Joisher
Jul 5, 2014

Derivate of function is slope of the tangent ... logy = e^-x On differentiation we get dy/dx = -(e^x)*y/ e^2x As x=0 y=e we get dy/dx = -e