Mathematical Induction (1 Viewer)

[kimmeh]

I got a question that i'm stuck on:

given that y = xe^x,
prove dy/dx = blah
prove d^2y/dx^2 = blah

then it says: predict the result for general solution for d^ny/dx^2 and prove by induction. <-How do i break that derivative up? it looks like a chain rule or something..

 

[Constip8edSkunk]

do you mean for d^n/dx^n?

y = xe^x
dy/dx = e^x(1+x)
d^2y/dx^x =e^x(2+x)
so prediction: d^ny/dx^n = e^x(n+x)
clearly true for n =1 see above
if d^ky/dx^k = e^x(k+x)
d^(k+1)y/dx^(k+1) = d/dx(d^ky/dx^k)
=d/dx(e^x(k+x))
=e^x+e^x(k+x)
=e^x((k+1)+x)
so by principle of induction d^ny/dx^n = e^x(n+x) is true

 

[withoutaface]

do you mean for d^n/dx^n?

y = xe^x
dy/dx = e^x(1+x)
d^2y/dx^x =e^x(2+x)
so prediction: d^ny/dx^n = e^x(n+x)
clearly true for n =1 see above
if d^ky/dx^k = e^x(k+x)
d^(k+1)y/dx^(k+1) = d/dx(d^ky/dx^k)
=d/dx(e^x(k+x))
=e^x+e^x(k+x)
=e^x((k+1)+x)
so by principle of induction d^ny/dx^n = e^x(n+x) is true

Yes she did.