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Find dy/dx for cube root of y - ln y = x + 8 using Implicit Differentiation
Mathematics
Grade 11 (Senior High School)
Question Content
Find $\frac{dy}{dx}$ for $\\sqrt[3]{y} - \\ln y = x + 8$
Correct Answer
$\\frac{dy}{dx} = \\frac{1}{\\frac{1}{3y^{2/3}} - \\frac{1}{y}}$ or simplified $\\frac{3y^{2/3}}{1 - 3y^{1/3}}$
Detailed Solution Steps
1
Step 1: Rewrite $\\sqrt[3]{y}$ as $y^{1/3}$. Differentiate both sides with respect to $x$: for the left side, differentiate $y^{1/3}$ to get $\\frac{1}{3}y^{-2/3}\\frac{dy}{dx}$ and $\\ln y$ to get $\\frac{1}{y}\\frac{dy}{dx}$; for the right side, differentiate $x+8$ to get $1$. This gives $\\frac{1}{3y^{2/3}}\\frac{dy}{dx} - \\frac{1}{y}\\frac{dy}{dx} = 1$.
2
Step 2: Factor out $\\frac{dy}{dx}$ from the left side: $\\frac{dy}{dx}\\left(\\frac{1}{3y^{2/3}} - \\frac{1}{y}\\right) = 1$.
3
Step 3: Solve for $\\frac{dy}{dx}$ by dividing both sides by $\\left(\\frac{1}{3y^{2/3}} - \\frac{1}{y}\\right)$: $\\frac{dy}{dx} = \\frac{1}{\\frac{1}{3y^{2/3}} - \\frac{1}{y}}$. To simplify, get a common denominator in the denominator: $\\frac{1}{\\frac{y - 3y^{2/3}}{3y^{5/3}}} = \\frac{3y^{5/3}}{y - 3y^{2/3}} = \\frac{3y^{2/3}}{1 - 3y^{1/3}}$.
Knowledge Points Involved
1
Implicit Differentiation
Used to differentiate equations with $y$ not isolated, treating $y$ as a function of $x$.
2
Derivative of Logarithmic Functions
The derivative of $\\ln y$ with respect to $x$ is $\\frac{1}{y}\\frac{dy}{dx}$, using the chain rule since $y$ is a function of $x$.
3
Power Rule for Rational Exponents
The derivative of $y^{1/3}$ with respect to $x$ is $\\frac{1}{3}y^{-2/3}\\frac{dy}{dx}$, applying the power rule $\\frac{d}{dx}(y^n) = ny^{n-1}\\frac{dy}{dx}$ for $n=1/3$.
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