How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation of $x^{3} + y^{3} = 4 x y + 1$ $x^3+y^3=4xy+1$ and evaluate at point (2,1)?

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How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation of $x^{3} + y^{3} = 4 x y + 1$ $x^3+y^3=4xy+1$ and evaluate at point (2,1)?

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Answer 1 · 2017-10-27T08:26:33

$\frac{8}{5}$ $8/5$

Explanation:

$\frac{d}{d x} (x^{3} + y^{3} = 4 x y + 1) - - (1)$ $d/(dx)(x^3+y^3=4xy+1)--(1)$

we will differentiate $w r t x$ $wrt" "x$ . Remembering that when differentiating $y$ $y$ we multiply by $\frac{d y}{d x}$ $(dy)/(dx)$ by virtue of the chain rule. Also on the $R H S$ $RHS$ we will need the product rule on the first term

$(1) \to 3 x^{2} + 3 y^{2} \frac{d y}{d x} = 4 y + 4 x \frac{d y}{d x} + 0$ $(1)rarr3x^2+3y^2(dy)/(dx)=4y+4x(dy)/(dx)+0$

rearrange for $\frac{d y}{d x}$ $(dy)/(dx)$

$3 y^{2} \frac{d y}{d x} - 4 x \frac{d y}{d x} = 4 y - 3 x^{2}$ $3y^2(dy)/(dx)-4x(dy)/(dx)=4y-3x^2$

$\frac{d y}{d x} (3 y^{2} - 4 x) = 4 y - 3 x^{2}$ $(dy)/(dx)(3y^2-4x)=4y-3x^2$

$\Rightarrow \frac{d y}{d x} = \frac{4 y - 3 x^{2}}{3 y^{2} - 4 x}$ $=>(dy)/(dx)=(4y-3x^2)/(3y^2-4x)$

$:.[(dy)/(dx)]_(color(white)(=)(2,1))=(4xx1-3xx2^2)/(3xx1^2-4xx2$

$=(4-12)/(3-8)=-8/-5=8/5$

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How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation of $x^{3} + y^{3} = 4 x y + 1$ $x^3+y^3=4xy+1$ and evaluate at point (2,1)?

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Explanation:

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How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation of $x^{3} + y^{3} = 4 x y + 1$ $x^3+y^3=4xy+1$ and evaluate at point (2,1)?