<s>[INST] Derivative Integral $$\frac{d}{dx} \int_{x}^{\sqrt{x}} \frac{e^{xy^2}}{y}dy$$
Where am I going wrong?
$$\begin{align*}
\frac{d}{dx} \int_{x}^{\sqrt{x}} \frac{e^{xy^2}}{y}dy = & \frac{d}{dx} ( F(\sqrt{x} ) - F(x))\\
= & \frac{F'(\sqrt{x} )}{2\sqrt{x}} - F'(x))\\
= & \frac{e^{x(\sqrt{x})^2}}{2\sqrt{x}\sqrt{x}} - \frac{e^{xx^2}}{x}\\
= & \frac{e^{x^2} - 2e^{x^3}}{2x}
\end{align*}$$
The alternatives to this question were
a) $\frac{e^{x^2}}{2x}$
b) $\frac{3e^{x^3}}{2x}$
c) $2e^{x^2} - 3e^{x^3}$
d) $\frac{ 2e^{x^2} - 3e^{x^3} }{2x}$
e) $\frac{ 2e^{x^2} + 3e^{x^3} }{2x}$
Thanks in advance!
 [/INST] We have from Leibniz's Rule for Differentiating Under the Integral
$$\begin{align}
\frac{d}{dx}\int_x^{\sqrt x}\frac{e^{xy^2}}{y}\,dy&=\color{blue}{\underbrace{\int_x^{\sqrt x}\frac{\partial}{\partial x}\left(\frac{e^{xy^2}}{y}\right)\,dy}_{\text{The Ommitted Term}}}+\color{red}{\underbrace{\left(\frac1{2\sqrt x}\frac{e^{x^2}}{\sqrt x}-\frac{e^{x^3}}{x}\right)}_{\text{This part was correct}}}\\\\
&=\color{blue}{\left(\frac{e^{x^2}}{2x}-\frac{e^{x^3}}{2x}\right)}+\color{red}{\left(\frac{e^{x^2}}{2 x}-\frac{e^{x^3}}{x}\right)}\\\\
&=\frac{2e^{x^2}-3e^{x^3}}{2x}
\end{align}$$
</s>