How do you differentiate #-y=cosy/x-x^2/y#?

Question

Aurora Alvarado · Accepted Answer

#y'=\frac{2x^3y+y^2\cos y}{x^4+x^2y^2-xy^2\sin y}#

Given equation:

#-y=\cos y/x-x^2/y#

#y=x^2/y-\cos y/x#

differentiating the above equation w.r.t. #x# on both sides as follows

#\frac{d}{dx}y=\frac{d}{dx}(x^2/y)-\frac{d}{dx}(\cos y/x)#

#y'=\frac{y\frac{d}{dx}x^2-x^2\frac{d}{dx}y}{y^2}-\frac{x\frac{d}{dx}\cos y-\cos y\frac{d}{dx}x}{x^2}#

#y'=\frac{2xy-x^2y'}{y^2}-\frac{-x\siny\ y'-\cos y}{x^2}#

#x^2y^2y'=2x^3y-x^4y'+xy^2\sin y\ y'+y^2\cos y#

#(x^4+x^2y^2-xy^2\sin y)y'=2x^3y+y^2\cos y#

#y'=\frac{2x^3y+y^2\cos y}{x^4+x^2y^2-xy^2\sin y}#

Cameron Alexander · Answer

undefined To differentiate the equation $ -y = \frac{\cos y}{x} - \frac{x^2}{y} $, you would use the quotient rule and the chain rule. Here's the step-by-step process:

1. Apply the quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$

2. Identify $u$ and $v$ in your equation:
   $u = \cos y$, $v = x$

3. Compute the derivatives of $u$ and $v$ with respect to $x$:
   $u' = -\sin y \frac{dy}{dx}$, $v' = 1$

4. Substitute these into the quotient rule formula and simplify:

$$
\frac{d}{dx}\left(\frac{\cos y}{x}\right) = \frac{-\sin y \frac{dy}{dx} \cdot x - \cos y}{x^2}
$$

5. Apply the quotient rule again for the second term:

$$
\frac{d}{dx}\left(-\frac{x^2}{y}\right) = \frac{-(2x)(y) - (-x^2)(\frac{dy}{dx})}{y^2}
$$

6. Combine the two derivatives:

$$
\frac{dy}{dx} = \frac{-\sin y \cdot x - \cos y - 2xy + x^2 \cdot \frac{dy}{dx}}{x^2 - 2xy}
$$

7. Solve for $\frac{dy}{dx}$ by isolating terms involving $\frac{dy}{dx}$ on one side:

$$
\frac{dy}{dx} - x^2 \cdot \frac{dy}{dx} = -\sin y \cdot x - \cos y - 2xy
$$

$$
\frac{dy}{dx}(1 - x^2) = -\sin y \cdot x - \cos y - 2xy
$$

$$
\frac{dy}{dx} = \frac{-\sin y \cdot x - \cos y - 2xy}{1 - x^2}
$$

This is the derivative of $y$ with respect to $x$.