Given: $$ x^{3}-2 x y+y^{3}=5 x $$, find $$ \frac{d y}{d x} $$ at $$ (1,2) 4 $$

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The value of $$\frac{dy}{dx}$$ at the point $$(1, 2)$$ is $$\frac{3}{5}$$. To find $$\frac{dy}{dx}$$ for the equation $$x^3 - 2xy + y^3 = 5x$$ at the point $$(1, 2)$$, we will use implicit differentiation. 1. Differentiate both sides of the equation with respect to $$x$$: $$ \frac{d}{dx}(x^3) - \frac{d}{dx}(2xy) + \frac{d}{dx}(y^3) = \frac{d}{dx}(5x) $$ 2. Applying the differentiation rules: - The derivative of $$x^3$$ is $$3x^2$$. - For $$2xy$$, we use the product rule: $$\frac{d}{dx}(2xy) = 2\left(x\frac{dy}{dx} + y\right)$$. - The derivative of $$y^3$$ is $$3y^2\frac{dy}{dx}$$. - The derivative of $$5x$$ is $$5$$. Putting it all together, we have: $$ 3x^2 - 2\left(x\frac{dy}{dx} + y\right) + 3y^2\frac{dy}{dx} = 5 $$ 3. Simplifying the equation: $$ 3x^2 - 2x\frac{dy}{dx} - 2y + 3y^2\frac{dy}{dx} = 5 $$ 4. Rearranging terms to isolate $$\frac{dy}{dx}$$: $$ (3y^2 - 2x)\frac{dy}{dx} = 5 - 3x^2 + 2y $$ 5. Solving for $$\frac{dy}{dx}$$: $$ \frac{dy}{dx} = \frac{5 - 3x^2 + 2y}{3y^2 - 2x} $$ 6. Now we will evaluate $$\frac{dy}{dx}$$ at the point $$(1, 2)$$: Substituting $$x = 1$$ and $$y = 2$$: $$ \frac{dy}{dx} = \frac{5 - 3(1)^2 + 2(2)}{3(2)^2 - 2(1)} $$ Calculating the numerator: $$ 5 - 3(1) + 4 = 5 - 3 + 4 = 6 $$ Calculating the denominator: $$ 3(4) - 2 = 12 - 2 = 10 $$ Thus, we have: $$ \frac{dy}{dx} = \frac{6}{10} = \frac{3}{5} $$ Therefore, the value of $$\frac{dy}{dx}$$ at the point $$(1, 2)$$ is $$ \boxed{\frac{3}{5}}. $$

Given: \( x^{3}-2 x y+y^{3}=5 x \), find \( \frac{d y}{d x} \) at \( (1,2) 4 \)

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Given: \( x^{3}-2 x y+y^{3}=5 x \), find \( \frac{d y}{d x} \) at \( (1,2) 4 \)

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