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

Question

Emma Aldridge · Accepted Answer

#ycos(x^2) - y^2 = y/x^2# #cos(x^2) = 1/x^2+y# #y = cos(x^2)-1/x^2# #y' = -2xsin(x^2) + 2/x^3#

Adam Adkins · Answer

undefined To differentiate the given equation, $y\cos(x^2) - y^2 = \frac{y}{x^2}$, with respect to $x$, we'll use the implicit differentiation method.

Differentiating each term separately and applying the chain rule where necessary, we get:

$$\begin{split}\frac{d}{dx}\left(y\cos(x^2)\right) - \frac{d}{dx}(y^2) &= \frac{d}{dx}\left(\frac{y}{x^2}\right) \
\Rightarrow y(-\sin(x^2) \cdot 2x) + \cos(x^2) \cdot \frac{dy}{dx} - 2y\frac{dy}{dx} &= \frac{d}{dx}\left(\frac{1}{x^2}\right) \cdot y + \frac{1}{x^2} \cdot \frac{dy}{dx} \
\Rightarrow -2xy\sin(x^2) + \cos(x^2) \cdot \frac{dy}{dx} - 2y\frac{dy}{dx} &= -\frac{2y}{x^3} + \frac{1}{x^2} \cdot \frac{dy}{dx} \
\Rightarrow \left(\cos(x^2) - \frac{1}{x^2}\right) \cdot \frac{dy}{dx} - 2y\frac{dy}{dx} &= -\frac{2y}{x^3} + 2xy\sin(x^2) \
\Rightarrow \left(\cos(x^2) - \frac{1}{x^2} - 2y\right) \cdot \frac{dy}{dx} &= -\frac{2y}{x^3} + 2xy\sin(x^2) \
\Rightarrow \frac{dy}{dx} &= \frac{-\frac{2y}{x^3} + 2xy\sin(x^2)}{\cos(x^2) - \frac{1}{x^2} - 2y}\end{split}$$