Question

1. -/1 points find y if y = ln(7x² + 3y²). y = 23. -/1 points find y if x^y = y^x. y =

Explanation:

Step1: Differentiate both sides for first - problem

Differentiate $y = \ln(7x^{2}+3y^{2})$ with respect to $x$. By the chain - rule, the derivative of the left - hand side is $y'$, and the derivative of the right - hand side is $\frac{14x + 6yy'}{7x^{2}+3y^{2}}$. So we have $y'=\frac{14x + 6yy'}{7x^{2}+3y^{2}}$.

Step2: Solve for $y'$ in first - problem

Multiply both sides by $7x^{2}+3y^{2}$: $(7x^{2}+3y^{2})y'=14x + 6yy'$. Rearrange terms: $(7x^{2}+3y^{2})y'-6yy'=14x$. Factor out $y'$: $y'(7x^{2}+3y^{2}-6y)=14x$. Then $y'=\frac{14x}{7x^{2}+3y^{2}-6y}$.

Step3: Differentiate both sides for second - problem

Take the natural logarithm of both sides of $x^{y}=y^{x}$, getting $y\ln x=x\ln y$. Differentiate both sides with respect to $x$ using the product rule. The left - hand side derivative is $y'\ln x+\frac{y}{x}$, and the right - hand side derivative is $\ln y+\frac{x}{y}y'$.

Step4: Solve for $y'$ in second - problem

$y'\ln x+\frac{y}{x}=\ln y+\frac{x}{y}y'$. Rearrange terms: $y'\ln x-\frac{x}{y}y'=\ln y - \frac{y}{x}$. Factor out $y'$: $y'(\ln x-\frac{x}{y})=\ln y - \frac{y}{x}$. Then $y'=\frac{\ln y-\frac{y}{x}}{\ln x - \frac{x}{y}}=\frac{y(x\ln y - y)}{x(y\ln x - x)}$.

Answer:

For $y = \ln(7x^{2}+3y^{2})$, $y'=\frac{14x}{7x^{2}+3y^{2}-6y}$
For $x^{y}=y^{x}$, $y'=\frac{y(x\ln y - y)}{x(y\ln x - x)}$