How do you find the derivative of #2e^(sqrtx)#?

Question

1804 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-26T06:27:27

#= e^(sqrtx) /(sqrtx)#

using the chain rule you know that #(alpha e^{f(x)} )'= alpha f'(x) e^{f(x)}#

or you can see for yourself by noting that if

#y = alpha e^{f(x)}#

then

#y/alpha = e^{f(x)}#

and

#ln(y/alpha) = f(x)#

so #(ln(y/alpha))' = f'(x)#

ie #1/ (y/alpha) (1/alpha) y' = f'(x)#

so #y' = (y f'(x) ) = alpha f'(x) e^{f(x)}#

here that means that

#(2e^(sqrtx))' = 2e^(sqrtx) * (sqrtx)'#

#= 2e^(sqrtx) * 1/2 1/(sqrtx)#

#= e^(sqrtx) /(sqrtx)#