3. differentiate implicitly to find y a) 2y+xy - 1 = 0 b) √y = 2y - x c) tan x = x² + y² d)x sin y = y² + 1

Explanation:

Step1: Differentiate each term of a)

Differentiate $2y+xy - 1=0$ term - by - term. The derivative of $2y$ with respect to $x$ is $2y'$ using the chain - rule. The derivative of $xy$ using the product rule $(uv)' = u'v+uv'$ where $u = x$ and $v = y$ is $y+xy'$, and the derivative of the constant $-1$ is $0$. So we have $2y'+y + xy'=0$.

Step2: Solve for $y'$ in a)

Group the terms with $y'$: $(2 + x)y'=-y$. Then $y'=-\frac{y}{x + 2}$.

Step3: Differentiate each term of b)

Rewrite $\sqrt{y}=2y - x$ as $y^{\frac{1}{2}}=2y - x$. Differentiate term - by - term. The derivative of $y^{\frac{1}{2}}$ with respect to $x$ is $\frac{1}{2}y^{-\frac{1}{2}}y'$ by the chain - rule. The derivative of $2y$ is $2y'$ and the derivative of $-x$ is $-1$. So $\frac{1}{2\sqrt{y}}y'=2y'-1$.

Step4: Solve for $y'$ in b)

Move the terms with $y'$ to one side: $\frac{1}{2\sqrt{y}}y'-2y'=-1$. Factor out $y'$: $y'(\frac{1}{2\sqrt{y}}-2)=-1$. Then $y'=\frac{-1}{\frac{1}{2\sqrt{y}}-2}=\frac{2\sqrt{y}}{4\sqrt{y}-1}$.

Step5: Differentiate each term of c)

Differentiate $\tan x=x^{2}+y^{2}$ term - by - term. The derivative of $\tan x$ is $\sec^{2}x$, the derivative of $x^{2}$ is $2x$, and the derivative of $y^{2}$ is $2yy'$. So $\sec^{2}x = 2x+2yy'$.

Step6: Solve for $y'$ in c)

Isolate the term with $y'$: $2yy'=\sec^{2}x - 2x$. Then $y'=\frac{\sec^{2}x - 2x}{2y}$.

Step7: Differentiate each term of d)

Differentiate $x\sin y=y^{2}+1$ using the product rule on the left - hand side. The derivative of $x\sin y$ is $\sin y+xy'\cos y$ (product rule with $u = x$ and $v=\sin y$), the derivative of $y^{2}$ is $2yy'$ and the derivative of the constant $1$ is $0$. So $\sin y+xy'\cos y=2yy'$.

Step8: Solve for $y'$ in d)

Move the terms with $y'$ to one side: $xy'\cos y-2yy'=-\sin y$. Factor out $y'$: $y'(x\cos y - 2y)=-\sin y$. Then $y'=\frac{\sin y}{2y - x\cos y}$.

Answer:

a) $y'=-\frac{y}{x + 2}$ b) $y'=\frac{2\sqrt{y}}{4\sqrt{y}-1}$ c) $y'=\frac{\sec^{2}x - 2x}{2y}$ d) $y'=\frac{\sin y}{2y - x\cos y}$