Find $\dfrac{dy}{dx}$(Implicit differentation + Quotient + Trig)

Question

Question :

1 Question: Find $\frac{d y}{d x}$ if $\sin \left(x^{2}+y^{2}\right)=\frac{\sin x}{\cos y}$

I Applied the quotient rule simultaneously with implicit differentiation to be left with $\frac{d y}{d x}=\frac{\cos (x) \cos (y)}{\cos ^{2}(y)-\sin (y) \sin (x)}$

I then thought I needed to replace $\cos (\mathrm{y})$ with $\frac{\sin x}{\sin \left(x^{2}+y^{2}\right)}$ but then found myself at a dead end whilst trying to replace $\sin (y),$ so i was wandering is this correct as the question does not specify the answer should be given in terms of $\mathrm{x} ?$ but again why would they provide me with $\sin \left(x^{2}+y^{2}\right)=\frac{\sin x}{\cos y}$

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Answer 1

You have a formula of the form $f(x,y)=g(x,y)$. Implicit differentiation means that you consider $y$ as function of $x$ and differentiate to obtain $$ f_x+f_y\,y'=g_x+g_y\,y',\text{ with }y'=\frac{dy}{dx}\text{ and } f_x=\frac{\partial f}{\partial x}\text{ etc.} $$ and rearranging $$ y'=-\frac{f_x-g_x}{f_y-g_y} $$

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For simplicity, rearrange your equation to be quotient-free.

Answer 2

if you have the constraint $$\sin(x^2+y^2) = \frac{sin x }{\cos y} \tag 1$$ and want difference it implicitly, we can the constraint into $$\cos y \sin(x^2 + y^2)=\sin x $$ which is easier to difference. you get $$-\sin (x^2 + y^2)\sin y\, dy + \cos y \cos(x^2 + y^2)(2xdx + 2ydy) = \cos x dx. $$ separating the $dx$ and $dy$ and diving out gives $$\frac{dy}{dx} = \frac{\cos x - 2x\cos y \cos(x^2 + y^2)}{2y\cos y \cos(x^2 + y^2)-\sin y\sin(x^2+y^2)} \text{ and } (1).$$

there are several equivalent answers here. you need to remember to carry the constraint $(1)$ along with any expression for $\frac{dy}{dx}.$