Evaluating $\frac{d}{dx} \int_{1}^{3x} \left(5\sin (t)-7e^{4t}+1\right)\,\mathrm dt$ 
$$\dfrac{d}{dx} \int_{1}^{3x} \left(5\sin (t)-7e^{4t}+1\right)\,\mathrm dt$$  

The answer I come up with is:  $5\sin(3x)(3)-7e^{4(3x)}(3)$, however this was not on the answer choice. What is the correct way to do this? Thanks. 

Let $$\begin{align}
f(x)&=\dfrac{d}{dx} \int_{1}^{3x} \left(5\sin (t)-7e^{4t}+1\right)\,\mathrm dt\\
&=\dfrac{d}{dx} \int_{0}^{3x}\left( 5\sin (t)-7e^{4t}+1\right)\,\mathrm dt-\dfrac{d}{dx} \int_0^{1}\left(5\sin (t)-7e^{4t}+1\right)\,\mathrm dt\\
&=\dfrac{d}{dx} \int_{0}^{3x}\left( 5\sin (t)-7e^{4t}+1\right)\,\mathrm dt+0\\
&=\left( 5\sin (3x)-7e^{4\cdot3x}+1\right)\cdot3\\
&=15\sin (3x)-21e^{12x}+3\\
\end{align}$$

$$\dfrac{d}{dx} \int_{1}^{3x} \left(5\sin (t)-7e^{4t}+1\right)\,\mathrm dt=15\sin (3x)-21e^{12x}+3$$

