Answer
$v^{\prime}(x)=\dfrac{2x \cdot 3^{x^2}[(x^2+1)\ln 3-1] }{(x^2+1)^2}$
Work Step by Step
We have: $v(x)=\dfrac{3^{x^2}}{ x^{2}+1}$
We differentiate both sides with respect to $x$.
$v^{\prime}(x)=\dfrac{d}{dx} [\dfrac{3^{x^2}}{ x^{2}+1}]$
Use rules: $\displaystyle \frac{d}{dx}[a^{u}]=a^u \ln (a) \frac{du}{dx}$ and $\displaystyle \frac{d}{dx}[x^{n}]=nx^{n-1}$
Now, $v^{\prime}(x)=\dfrac{(x^2+1)(3^{x^2})(\ln 3) (2x) -3^{x^2}(2x)}{(x^2+1)^2}$
So,
$v^{\prime}(x)=\dfrac{2x \cdot 3^{x^2}[(x^2+1)\ln 3-1] }{(x^2+1)^2}$
