How do you find the slope of the tangent line to the curve $x y^{3} + x y = 4$ $xy^3 + xy = 4$ at the point (2,1)? Thank you! I really, really, appreciate your help:)?

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How do you find the slope of the tangent line to the curve $x y^{3} + x y = 4$ $xy^3 + xy = 4$ at the point (2,1)? Thank you! I really, really, appreciate your help:)?

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Answer 1 · 2015-09-26T02:17:26

$s l o p e = - \frac{1}{4}$ $slope =-1/4$

Explanation:

$x y^{3} + x y = 4$ $xy^3+xy=4$
$y^{3} + 3 x y^{2} \frac{d y}{d x} + y + x \frac{d y}{d x} = 0$ $y^3+3xy^2dy/dx+y+xdy/dx=0$
slope at (2,1)
$1 + 6 \frac{d y}{d x} + 1 + 2 \frac{d y}{d x} = 0$ $1+6dy/dx+1+2dy/dx=0$
$8 \frac{d y}{d x} = - 2$ $8dy/dx=-2$
$\frac{d y}{d x} = - \frac{1}{4} \Rightarrow s l o p e = - \frac{1}{4}$ $dy/dx=-1/4=>slope =-1/4$

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How do you find the slope of the tangent line to the curve $x y^{3} + x y = 4$ $xy^3 + xy = 4$ at the point (2,1)? Thank you! I really, really, appreciate your help:)?

1 Answer

Explanation:

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How do you find the slope of the tangent line to the curve $x y^{3} + x y = 4$ $xy^3 + xy = 4$ at the point (2,1)? Thank you! I really, really, appreciate your help:)?