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

1 Answer
Mar 7, 2018

#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