Question

How do I find dx/dy by implicit differentiation and evaluate the derivative by the given point?
x cos y = 1 ; (2, π/3)

Answer 1

Since we'll be taking our derivative with respect to x, each time we differentiate a `non-x term`, we tack on a y'.\[\Large x \cos y = 1\]

On the left side we'll apply the `product rule`:\[\Large \color{royalblue}{(x)'}\cos y+x \color{royalblue}{(\cos y)'}=\color{royalblue}{(1)'}\]

Answer 2

So we need to take the derivative of the blue stuff.

Answer 3

derivative of 1?

Answer 4

The derivative of 1 is 0

Answer 5

But how do I evaluate the derivative at the point (2, π/3)

Answer 6

1 gives you 0? sounds right:
\[\Large \color{royalblue}{(x)'}\cos y+x \color{royalblue}{(\cos y)'}=\color{orangered}{0}\]

Understand how we calculated the left side?\[\Large \color{orangered}{(1)}\cos y+x \color{orangered}{(-\sin y)y'}=\color{orangered}{0}\]

Answer 7

Solve for y':\[\Large y'\quad=\quad \frac{\cos x}{x \sin y}\]See how the right side involves some x and y stuff? We can think of y' as a function of both x and y.

\[\Large y'(x,y)\quad=\quad \frac{\cos x}{x \sin y}\]And from here they want you to evaluate the derivative at x=2, y=pi/3.\[\Large y'\left(2,\;\frac{\pi}{3}\right)\quad=\quad?\]