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

Sep 26, 2014

We begin this problem by finding the point of tangency.

Substitute in the value of 1 for $x$.

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

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

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

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

${y}^{3 \cdot \frac{1}{3}} = {8}^{\frac{1}{3}}$

${y}^{\frac{3}{3}} = {8}^{\frac{1}{3}}$

${y}^{1} = {8}^{\frac{1}{3}}$

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

$y = {2}^{3 \cdot \frac{1}{3}}$

$y = {2}^{\frac{3}{3}}$

$y = {2}^{1}$

$y = 2$

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

Complete the Implicit Differentiation

$3 {x}^{2} + 3 {y}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

Substitute in those $x \mathmr{and} y$ values from above $\implies \left(1 , 2\right)$

$3 {\left(1\right)}^{2} + 3 {\left(2\right)}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$3 + 3 \cdot 4 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$3 + 12 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$12 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - 3$

$\frac{12 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{12} = \frac{- 3}{12}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{4} = - 0.25 \implies S l o p e = m$

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

We have $\left(x , y\right) \implies \left(1 , 2\right)$

We have $m = - 0.25$

Make the substitutions

$y = m x + b$

$2 = - 0.25 \left(1\right) + b$

$2 = - 0.25 + b$

$0.25 + 2 = b$

$2.25 = b$

Equation of the tangent line ...

$y = - 0.25 x + 2.25$

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

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