Explanation

(d)
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 $\frac{dv}{dx\ } = e^{2x - 1}(2)$
$\therefore\ \frac{du}{dv} = \frac{e^{x^{2}}.2x}{e^{2x - 1}.2}$
$\Rightarrow \frac{du}{dv} = xe^{x^{2} - 2x + 1}$
$\Rightarrow \ \ \ \left( \frac{du}{dv} \right)_{(x = 1)} = \$1.$e^{1 - 2 + 1} = 1\$