Integration and derivative I want to compute the derivative
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{x^3}^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y\right).
$$
Can anyone give me some hints for this question? It is too complex and I don't know what is the first step I can do? Thank you.
 A: Hint: The Leibniz Integral rule states that: $$\frac{\mathrm d}{\mathrm dx}\int_{f_{1}(x)}^{f_{2}(x)}g(t)\mathrm dt=g(f_{2}(x))f_{2}'(x)-g(f_{1}(x))f_{1}'(x)$$

Here the expression simplifies to:

 $$\frac{2e^{2x}\sin(3e^{2x})}{\sqrt{e^{4x}+\cos(3e^{2x})+1}}-\frac{3x^{2}\sin(3x^{3})}{\sqrt{x^{6}+\cos(3x^{3})+1}}$$

A: Use Leibniz integral rule for differentiation under the integral sign.
A: Notice that we can split the integral up into
\begin{align*}
\int_{x^3}^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y
&=\int_a^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y+\int_{x^3}^a\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y \\
&=\int_a^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y-\int_a^{x^3}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y
\end{align*}
for some $a\in\mathbb{R}$. Applying the chain rule and the fundamental theorem of calculus we thus get that
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{x^3}^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y\right)
&=\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_a^{e^{2x}}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y\right)-\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_a^{x^3}\frac{\sin(3y)}{\sqrt{y^2+\cos(3y)+1}}\,\mathrm{d}y\right) \\
&=\frac{2e^{2x}\sin(3e^{2x})}{\sqrt{e^{4x}+\cos(3e^{2x})+1}}-\frac{3x^2\sin(3x^3)}{\sqrt{x^6+\cos(3x^3)+1}}
\end{align*}
A: Alternative to ganesh's suggestion:
Let $a\in\mathbb R.$ Then $$\frac{\mathrm d}{\mathrm dx}\int_{x^3}^{e^{2x}}\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}\,\mathrm dy\stackrel{\text{by convention}}{=}\frac{\mathrm d}{\mathrm dx}\left(\int_{a}^{e^{2x}}\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}\,\mathrm dy-\int_{a}^{x^3}\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}\,\mathrm dy\right)\\
=\frac{\mathrm d}{\mathrm dx}\int_{a}^{e^{2x}}\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}\,\mathrm dy-\frac{\mathrm d}{\mathrm dx}\int_{a}^{x^3}\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}\,\mathrm dy\\
=\frac{2e^{2x}\sin\left(3e^{2x}\right)}{\sqrt{e^{4x}+\cos\left(3e^{2x}\right)+1}}-\frac{3x^2\sin\left(3x^3\right)}{\sqrt{x^{6}+\cos\left(3x^3\right)+1}}$$ (by the chain rule and the first fundamental theorem of calculus, since $\dfrac{\sin\left(3y\right)}{\sqrt{y^{2}+\cos\left(3y\right)+1}}$ is continuous).
In general, if $g$ is differentiable and $f$ is continuous on the domain of integration below, then $$\frac{\mathrm d}{\mathrm dx}\int_a^{g(x)}f(t)\,\mathrm dt=f\Big(g(x)\Big)g'(x).$$
