Question

Find the equation of tangent at point where the curve y=2e^(-x/3)?

Answer 1

`y'=-2/3e^(-x/3)`

Explanation:

`y= 2e^(-x/3)`

This is the general rule when dealing with the irrational number `e`

If `y=e^(f(x))`, then `y'=f'(x)e^(f(x))`

So, the differential is

`y'=-2/3e^(-x/3)`

To find an equation of a tangent at a point on this graph, we would need a point, which is not given in the question