미적분의 기본정리

(a) Using the product rule, followed by the chain rule, followed by the Fundamental Theorem of Calculus Part 1,

$$\begin{aligned} f'(x) & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\frac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{x^3+1} e^{t^3} \mathrm{d}t \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\ \big[3x^2\big] \Bigg[\frac{d}{\mathrm{d}y}\int_{0}^{y} e^{t^3} \mathrm{d}t\Bigg]_{y=x^3+1} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\ \big[3x^2\big] \Big[e^{y^3}\Big]_{y=x^3+1} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + x^3\ \big[3x^2\big] e^{(x^3+1)^3} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \mathrm{d}t + 3x^5 e^{(x^3+1)^3} \end{aligned}$$

(b) In general, the equation of the tangent line to the graph of $y=f(x)$ at $x=a$ is

$$y=f(a) + f'(a)\,(x-a)$$

Substituting in the given $f(x)$ and $a=-1$:

$$\begin{aligned} f(a)=f(-1)&=(-1)^3\int_0^0 e^{t^3}\ \mathrm{d}t=0\\ f'(a)=f'(-1)&=3(-1)^2\int_{0}^{0}e^{t^3}\ \mathrm{d}t + 3(-1)^5e^{0}\\ &=0-3=-3\\ (x-a)=x-(-1)&=x+1 \\ \text{So, the equation of the tangent line is} \\ y = -3(x+1)\ . \end{aligned}$$