How do you Use implicit differentiation to find the equation of the tangent line to the curve $x^{3} + y^{3} = 9$ $x^3+y^3=9$ at the point where $x = - 1$ $x=-1$ ?

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How do you Use implicit differentiation to find the equation of the tangent line to the curve $x^{3} + y^{3} = 9$ $x^3+y^3=9$ at the point where $x = - 1$ $x=-1$ ?

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Answer 1 · 2014-09-26T01:23:48

We begin this problem by finding the point of tangency.

Substitute in the value of 1 for $x$ $x$ .

$x^{3} + y^{3} = 9$ $x^3+y^3=9$
${(1)}^{3} + y^{3} = 9$ $(1)^3+y^3=9$
$1 + y^{3} = 9$ $1+y^3=9$
$y^{3} = 8$ $y^3=8$

Not sure how to show a cubed root using our math notation here on Socratic but remember that raising a quantity to the $\frac{1}{3}$ $1/3$ power is equivalent.

Raise both sides to the $\frac{1}{3}$ $1/3$ power

${(y^{3})}^{\frac{1}{3}} = 8^{\frac{1}{3}}$ $(y^3)^(1/3)=8^(1/3)$

$y^{3 \cdot \frac{1}{3}} = 8^{\frac{1}{3}}$ $y^(3*1/3)=8^(1/3)$

$y^{\frac{3}{3}} = 8^{\frac{1}{3}}$ $y^(3/3)=8^(1/3)$

$y^{1} = 8^{\frac{1}{3}}$ $y^(1)=8^(1/3)$

$y = {(2^{3})}^{\frac{1}{3}}$ $y=(2^3)^(1/3)$

$y = 2^{3 \cdot \frac{1}{3}}$ $y=2^(3*1/3)$

$y = 2^{\frac{3}{3}}$ $y=2^(3/3)$

$y = 2^{1}$ $y=2^(1)$

$y = 2$ $y=2$

We just found that when $x = 1, y = 2$ $x=1, y=2$

Complete the Implicit Differentiation

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

Substitute in those $x and y$ $x and y$ values from above $\Rightarrow (1, 2)$ $=>(1,2)$

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

$3 + 3 \cdot 4 (\frac{d y}{d x}) = 0$ $3+3*4(dy/dx)=0$

$3 + 12 (\frac{d y}{d x}) = 0$ $3+12(dy/dx)=0$

$12 (\frac{d y}{d x}) = - 3$ $12(dy/dx)=-3$

$\frac{12 (\frac{d y}{d x})}{12} = \frac{- 3}{12}$ $(12(dy/dx))/12=(-3)/12$

$\frac{d y}{d x} = \frac{- 1}{4} = - 0.25 \Rightarrow S l o p e = m$ $(dy)/dx=(-1)/4=-0.25 => Slope = m$

Now use the slope intercept formula, $y = m x + b$ $y=mx+b$

We have $(x, y) \Rightarrow (1, 2)$ $(x,y) => (1,2)$

We have $m = - 0.25$ $m = -0.25$

Make the substitutions

$y = m x + b$ $y=mx+b$

$2 = - 0.25 (1) + b$ $2 = -0.25(1)+b$

$2 = - 0.25 + b$ $2 = -0.25+b$

$0.25 + 2 = b$ $0.25 + 2=b$

$2.25 = b$ $2.25=b$

Equation of the tangent line ...

$y = - 0.25 x + 2.25$ $y=-0.25x+2.25$

To get a visual with the calculator solve the original equation for $y$ $y$ .

$y = {(9 - x^{3})}^{\frac{1}{3}}$ $y=(9-x^3)^(1/3)$

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How do you Use implicit differentiation to find the equation of the tangent line to the curve $x^{3} + y^{3} = 9$ $x^3+y^3=9$ at the point where $x = - 1$ $x=-1$ ?

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How do you Use implicit differentiation to find the equation of the tangent line to the curve $x^{3} + y^{3} = 9$ $x^3+y^3=9$ at the point where $x = - 1$ $x=-1$ ?