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Find dy/dx for 6 - xe^(y^5) = 2x^10 using Implicit Differentiation
Mathematics
Grade 11 (Senior High School)
Question Content
Find $\frac{dy}{dx}$ for $6 - xe^{y^5} = 2x^{10}$
Correct Answer
$\\frac{dy}{dx} = \\frac{20x^9 + e^{y^5}}{5xy^4e^{y^5}}$
Detailed Solution Steps
1
Step 1: Differentiate both sides with respect to $x$. Differentiate $6$ to get $0$, use the product rule on $-xe^{y^5}$: $-\\left(e^{y^5} + x \\cdot e^{y^5} \\cdot 5y^4\\frac{dy}{dx}\\right)$, and differentiate $2x^{10}$ to get $20x^9$. This gives $0 - e^{y^5} - 5xy^4e^{y^5}\\frac{dy}{dx} = 20x^9$.
2
Step 2: Rearrange to isolate the term with $\\frac{dy}{dx}$: $-5xy^4e^{y^5}\\frac{dy}{dx} = 20x^9 + e^{y^5}$.
3
Step 3: Divide both sides by $-5xy^4e^{y^5}$ to solve for $\\frac{dy}{dx}$: $\\frac{dy}{dx} = -\\frac{20x^9 + e^{y^5}}{-5xy^4e^{y^5}} = \\frac{20x^9 + e^{y^5}}{5xy^4e^{y^5}}$.
Knowledge Points Involved
1
Implicit Differentiation
Used to differentiate equations with $y$ inside a composite function, treating $y$ as a function of $x$ and applying chain and product rules.
2
Product Rule for Differentiation
States that $\\frac{d}{dx}(uv) = u'v + uv'$, where $u$ and $v$ are functions of $x$. Used here to differentiate $xe^{y^5}$, with $u=x$ and $v=e^{y^5}$.
3
Chain Rule for Exponential Functions
The derivative of $e^{u}$ with respect to $x$ is $e^{u}\\frac{du}{dx}$, where $u$ is a function of $x$. Here, $u=y^5$, so the derivative is $e^{y^5} \\cdot 5y^4\\frac{dy}{dx}$.
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