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

1 Answer
Youssuf E.
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

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