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

Question

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

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.

Answer 1

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

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

Answer 2

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

use this $$\dfrac{d}{dx}\int_{u(x)}^{v(x)}f(t)dt=f(u(x))\cdot u'(x)-f(v(x))\cdot v'(x)$$