QuestionExercises - find the derivative of $y=f(x)$ at $x=0$ where $y$ is determined by $y^{5}+2 y-x-3 x^{7}=0$ - find the tangent line of $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ at $\left(2, \frac{3 \sqrt{3}}{2}\right)$

Studdy Solution
Use the point-slope form of a line $y - y_1 = m(x - x_1)$ to find the equation of the tangent line, where $m = -\frac{\sqrt{3}}{4}$ and the point is $\left(2, \frac{3\sqrt{3}}{2}\right)$:
$$y - \frac{3\sqrt{3}}{2} = -\frac{\sqrt{3}}{4}(x - 2)$$
Simplify to get the equation of the tangent line:
$$y = -\frac{\sqrt{3}}{4}x + \frac{\sqrt{3}}{2} + \frac{3\sqrt{3}}{2}$$
$$y = -\frac{\sqrt{3}}{4}x + 2\sqrt{3}$$
The derivative of $y$ at $x = 0$ is $\frac{1}{2}$, and the equation of the tangent line is $y = -\frac{\sqrt{3}}{4}x + 2\sqrt{3}$.