[tex]y=(x^3+7)^4(x^2-3)^3[/tex]
Let's find the derivative of this curve:
[tex]\frac{dy}{dx}=12x^2(x^3+7)^3(x^2-3)^3+6x(x^2-3)^2(x^3+7)^4[/tex]
Evaluate it for x = -2:
[tex]\frac{dy}{dx}\begin{cases} \\ x=-2\end{cases}=12(4)(-1)(1)+6(-2)(1)(1)=-48-12=-60[/tex]
Let:
m1 = -60
Since the line is normal, the other line, must have a slope m2 of the form:
[tex]\begin{gathered} m1\cdot m2=-1 \\ m2=-\frac{1}{m1} \\ m2=-\frac{1}{-60} \\ m2=\frac{1}{60} \end{gathered}[/tex]
Using the point-slope equation:
[tex]\begin{gathered} y-y1=m(x-x1) \\ _{\text{ }}where\colon \\ y1=y(-2)=(1)(1)=1 \\ so\colon \\ y-1=\frac{1}{60}(x+2) \\ y=\frac{1}{60}x+\frac{31}{30} \end{gathered}[/tex]
Rewrite the equation as: Ax + By + C = 0
[tex]\begin{gathered} \frac{1}{60}x-y+\frac{31}{30}=0 \\ x-60y+62=0 \end{gathered}[/tex]
Answer:
x - 60y + 62 = 0
