The value of differentiation of ex2 with respect toe2x–1 at x=1 is
e
0
e-1
1
Explanation for the correct option:
Find the required value.
Letu=ex2
On differentiating w.r.t. x, we get
dudx=ex22x
Let v=e2x–1
dvdx=2e2x–1
∴dudv=dudxdvdx
=ex22x2e2x–1=xex2-2x+1
Now, dudvx=1=1×e1-2+1
=1
Hence, option (D) is the correct answer.