4. use implicit differentiation to find y for x² - √cosy + y² = 1

Explanation:

Step1: Differentiate each term

Differentiate $x^{2}$ with respect to $x$ gives $2x$. For $-\sqrt{\cos y}$, use the chain - rule. Let $u = \cos y$, then $-\sqrt{u}= - u^{\frac{1}{2}}$. The derivative of $-u^{\frac{1}{2}}$ with respect to $u$ is $-\frac{1}{2}u^{-\frac{1}{2}}$, and the derivative of $u=\cos y$ with respect to $x$ is $-\sin y\cdot y'$. So the derivative of $-\sqrt{\cos y}$ with respect to $x$ is $\frac{\sin y\cdot y'}{2\sqrt{\cos y}}$. The derivative of $y^{2}$ with respect to $x$ using the chain - rule is $2y\cdot y'$, and the derivative of the right - hand side $1$ with respect to $x$ is $0$. So we have: $2x+\frac{\sin y\cdot y'}{2\sqrt{\cos y}} + 2y\cdot y'=0$

Step2: Isolate $y'$

First, move the non - $y'$ term to the other side: $\frac{\sin y\cdot y'}{2\sqrt{\cos y}}+2y\cdot y'=- 2x$ Factor out $y'$: $y'\left(\frac{\sin y}{2\sqrt{\cos y}} + 2y\right)=-2x$ Then solve for $y'$: $y'=\frac{-2x}{\frac{\sin y}{2\sqrt{\cos y}}+2y}=\frac{-4x\sqrt{\cos y}}{\sin y + 4y\sqrt{\cos y}}$

Answer:

$y'=\frac{-4x\sqrt{\cos y}}{\sin y + 4y\sqrt{\cos y}}$