Respuesta :
Use the product rule!
f(x) = (x^2)(e^5x)
First times the derivative of the second, plus, second times the derivative of the first:
(d/dx) f(x) = (x^2) d/dx(e^5x) + (e^5x) d/dx (x^2)
what's the derivative of e^5x? well you have to chain rule for that:
d/dx (e^5x) = (e^5x) (5)
now what is the derivative of x^2? yup! just 2
d/dx (x^2) = 2
so what's f'(x)? Let's just plug in our derivatives
f'(x) = (x^2) (e^5x) (5) + (e^5x)(2)
tada!
f ` ( x ) = ( x² )` · e^(5x) + x² · ( e^(5x) )` =
= 2 x · e^(5x) + 5 e^(5x) · x² =
= x e^(5x) ( 2 + 5 x )
f `` ( x ) = ( 2 x e^(5x) + 5 x² e^(5x) ) ` =
= ( 2 x ) ˙e^(5x) + 2 x ( e^(5x) )` + ( 5 x² ) ` · e^(5x) + ( e^(5x)) ` · 5 x² =
= 2 · e^(5x) + 10 x · e^(5x) + 10 x · e^(5x) + 25 x² · e^(5x) =
= e^(5x) · ( 2 + 20 x + 25 x² )
= 2 x · e^(5x) + 5 e^(5x) · x² =
= x e^(5x) ( 2 + 5 x )
f `` ( x ) = ( 2 x e^(5x) + 5 x² e^(5x) ) ` =
= ( 2 x ) ˙e^(5x) + 2 x ( e^(5x) )` + ( 5 x² ) ` · e^(5x) + ( e^(5x)) ` · 5 x² =
= 2 · e^(5x) + 10 x · e^(5x) + 10 x · e^(5x) + 25 x² · e^(5x) =
= e^(5x) · ( 2 + 20 x + 25 x² )