A) e
B) 0
C) \[{{e}^{-1}}\]
D) 1
Correct Answer: D
Solution :
Let\[u={{e}^{{{x}^{2}}}}\]and \[v={{e}^{2x-1}}\] On differentiating w.r.t.\[x,\]we get \[\frac{du}{dx}={{e}^{{{x}^{2}}}}.2x\] and \[\frac{dv}{dx}={{e}^{2x-1}}(2)\] \[\therefore \] \[\frac{du}{dv}=\frac{{{e}^{{{x}^{2}}}}.2x}{{{e}^{2x-1}}.2}\] \[\Rightarrow \] \[\frac{du}{dv}=x{{e}^{{{x}^{2}}-2x+1}}\] \[\Rightarrow \] \[{{\left( \frac{du}{dv} \right)}_{x=1}}=1.{{e}^{1-2+1}}\] \[\Rightarrow \] \[\frac{du}{dv}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
