How do you find #dy/dx# by implicit differentiation of #coty=x-y#?

Question

Isaiah Allen · Accepted Answer

We start by defining the derivative of #coty#. Recall that #cotx = 1/tanx = 1/(sinx/cosx) = cosx/sinx#.

#:.coty = cosy/siny#

#:.(coty)' = (-siny xx siny(dy/dx) - (cosy xx cosy)dy/dx)/(siny)^2#

#:.(coty)' = (-sin^2y(dy/dx) - cos^2y(dy/dx))/sin^2y#

#:.(coty)' = -(sin^2y(dy/dx) + cos^2y(dy/dx))/sin^2y#

#:.(coty)' =- (dy/dx)/sin^2y#

#:.(coty)' = -dy/dxcsc^2y#

Now, the rest of the relation.

#-csc^2y(dy/dx) = 1 - 1(dy/dx)#

#-csc^2y(dy/dx) + 1(dy/dx) = 1#

#dy/dx(-csc^2y + 1) = 1#

#dy/dx = 1/(1 - csc^2y)#

From the relation #1+cot^2y=csc^2y# we see that #1-csc^2y=-cot^2y#. We can rewrite the derivative if we wish.

#dy/dx=-1/cot^2y#

If we want, we can use the original function #coty=x-y#.

#dy/dx=-1/(x-y)^2#

Hopefully this helps!

Paisley Johnson · Answer

undefined To find $ \frac{dy}{dx} $ by implicit differentiation of $ \cot(y) = x - y $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Use the chain rule and derivative rules for trigonometric functions where necessary.
3. Solve for $ \frac{dy}{dx} $ after differentiation.

Here are the steps:

1. Differentiate $ \cot(y) $ with respect to $ x $ using the chain rule:
   $$ \frac{d}{dx}(\cot(y)) = \frac{d}{dx}(x - y) $$

2. Apply the chain rule to $ \cot(y) $:
   $$ -\csc^2(y) \frac{dy}{dx} = 1 - \frac{dy}{dx} $$

3. Rearrange the equation to isolate $ \frac{dy}{dx} $:
   $$ -\csc^2(y) \frac{dy}{dx} + \frac{dy}{dx} = 1 $$
   $$ \frac{dy}{dx} (-\csc^2(y) + 1) = 1 $$

4. Solve for $ \frac{dy}{dx} $:
   $$ \frac{dy}{dx} = \frac{1}{-\csc^2(y) + 1} $$

So, $ \frac{dy}{dx} = \frac{1}{-\csc^2(y) + 1} $ is the derivative of $ y $ with respect to $ x $ implicitly defined by the equation $ \cot(y) = x - y $.