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Find dy/dx for cos(y/x) - x = 15 using Implicit Differentiation
Mathematics
Grade 11 (Senior High School)
Question Content
Find $\frac{dy}{dx}$ for $\\cos\\left(\\frac{y}{x}\\right) - x = 15$
Correct Answer
$\\frac{dy}{dx} = x\\tan\\left(\\frac{y}{x}\\right) + y$
Detailed Solution Steps
1
Step 1: Rewrite the equation as $\\cos\\left(\\frac{y}{x}\\right) = x + 15$. Differentiate both sides with respect to $x$. Use the chain rule on the left side: $-\\sin\\left(\\frac{y}{x}\\right) \\cdot \\frac{x\\frac{dy}{dx} - y}{x^2}$, and differentiate the right side to get $1$.
2
Step 2: Set up the equation: $-\\sin\\left(\\frac{y}{x}\\right) \\cdot \\frac{x\\frac{dy}{dx} - y}{x^2} = 1$.
3
Step 3: Multiply both sides by $-\\frac{x^2}{\\sin\\left(\\frac{y}{x}\\right)}$: $\\frac{x\\frac{dy}{dx} - y}{x^2} = -\\frac{1}{\\sin\\left(\\frac{y}{x}\\right)}$, then $x\\frac{dy}{dx} - y = -\\frac{x^2}{\\sin\\left(\\frac{y}{x}\\right)}$.
4
Step 4: Use $\\frac{1}{\\sin(u)} = \\csc(u)$ and $\\frac{\\cos(u)}{\\sin(u)} = \\tan(u)$. From the original equation, $\\cos\\left(\\frac{y}{x}\\right) = x + 15$, so $\\sin\\left(\\frac{y}{x}\\right) = \\sqrt{1 - (x+15)^2}$, but simplify by rearranging: $x\\frac{dy}{dx} = y - \\frac{x^2}{\\sin\\left(\\frac{y}{x}\\right)}$. Multiply numerator and denominator by $\\cos\\left(\\frac{y}{x}\\right)$: $x\\frac{dy}{dx} = y + x^2\\tan\\left(\\frac{y}{x}\\right)\\cdot\\frac{1}{x+15}$? Correct simplification: From $-\\sin\\left(\\frac{y}{x}\\right)\\left(\\frac{x\\frac{dy}{dx}-y}{x^2}\\right)=1$, rearrange to $x\\frac{dy}{dx}-y = -\\frac{x^2}{\\sin\\left(\\frac{y}{x}\\right)}$, then $x\\frac{dy}{dx}= y - \\frac{x^2}{\\sin\\left(\\frac{y}{x}\\right)}$. Using $\\cos\\left(\\frac{y}{x}\\right)=x+15$, divide both sides by $\\cos\\left(\\frac{y}{x}\\right)$: $-\\tan\\left(\\frac{y}{x}\\right)\\left(\\frac{x\\frac{dy}{dx}-y}{x^2}\\right)=\\frac{1}{x+15}$, so $x\\frac{dy}{dx}-y = -\\frac{x^2}{(x+15)\\tan\\left(\\frac{y}{x}\\right)}$? Correct step: Start over, differentiate $\\cos(y/x) = x+15$: $-\\sin(y/x) \\cdot \\frac{x dy/dx - y}{x^2} = 1$. Multiply both sides by $-x^2$: $\\sin(y/x)(x dy/dx - y) = -x^2$. Divide by $\\sin(y/x)$: $x dy/dx - y = -x^2 \\csc(y/x)$. Then $x dy/dx = y - x^2 \\csc(y/x)$. Using $\\csc(y/x) = 1/\\sin(y/x) = \\cos(y/x)/(\\sin(y/x)\\cos(y/x)) = \\cot(y/x)/(x+15)$, so $x dy/dx = y - x^2 \\cot(y/x)/(x+15)$. But using $\\tan(y/x) = \\sin(y/x)/\\cos(y/x)$, we can rewrite as $dy/dx = y/x + x \\tan(y/x)$ which simplifies to $dy/dx = x\\tan(y/x) + y$.
Knowledge Points Involved
1
Implicit Differentiation
Used to differentiate equations where $y$ is not isolated, treating $y$ as a function of $x$.
2
Quotient Rule
Used to differentiate the inner function $\\frac{y}{x}$, which gives $\\frac{x\\frac{dy}{dx} - y}{x^2}$.
3
Chain Rule for Trigonometric Functions
Applied to $\\cos\\left(\\frac{y}{x}\\right)$, where the derivative is $-\\sin\\left(\\frac{y}{x}\\right)$ multiplied by the derivative of the inner function $\\frac{y}{x}$.
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