how do i find the derivative of 2xe^x^2?

Question

I have an equation based on the principle

  _lim     f(x)            _lim     f'(x)

^x→x₀  g(x)    ^{=    x→x₀}  g'(x)

 

 _lim     sin²x   _{=    lim}     2sinxcosx   ₌        cos2x l_{x=0      =}   1

^x→0    e^x²-1        ^x→0       2xe^x²             2e^x²(2x²+1) l_x=0      6

 

I don't understand how the transition is made from the second step to the third.
Do I say that f(x)=2x and g(x)=e^x²? how do i get g'(x)??
Thanks!

Answer 1

The second part of the question

lim   (sin²x)/(exp(x²)-1) using L'Hopital's rule
x→0

 

=lim(2sinxcosx)/(2xexp(x²))=lim  (sin2x/2x)(1/exp(x²)

  x→0                                    x→0

=lim (sin2x/2x)×lim(1/exp(x²)=1×1=1                                 

  x→0                 x→0

Answer 2

You use the product rule. d(f•g)/dx = g(df/dx) + f(dg/dx)

f = 2x

g = e^{x^2}

df/dx = 2

You use the chain rule for g. dg/dx = dg/dy(dydx) y = x²

d(e^y)dy = e^y dy/dx = 2x so dg/dx = 2xe^{x^2}

 

So d(2xe^{x^2})dx = 2e^{x^2} +2x(2xe^{x^2}) = 4x²e^{x^2}+2e^{x^2} = (4x²+2)e^{x^2}