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# ?

Question

Charles Anctil · Accepted Answer

We begin this problem by finding the point of tangency. Substitute in the value of 1 for #x#. #x^3+y^3=9#
#(1)^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 #1/3# power is equivalent. Raise both sides to the #1/3# power #(y^3)^(1/3)=8^(1/3)# #y^(3*1/3)=8^(1/3)# #y^(3/3)=8^(1/3)# #y^(1)=8^(1/3)# #y=(2^3)^(1/3)# #y=2^(3*1/3)# #y=2^(3/3)# #y=2^(1)# #y=2# We just found that when #x=1, y=2# Complete the Implicit Differentiation #3x^2+3y^2(dy/dx)=0# Substitute in those #x and y# values from above #=>(1,2)# #3(1)^2+3(2)^2(dy/dx)=0# #3+3*4(dy/dx)=0# #3+12(dy/dx)=0# #12(dy/dx)=-3# #(12(dy/dx))/12=(-3)/12# #(dy)/dx=(-1)/4=-0.25 => Slope = m# Now use the slope intercept formula, #y=mx+b# We have #(x,y) => (1,2)# We have #m = -0.25# Make the substitutions #y=mx+b# #2 = -0.25(1)+b# #2 = -0.25+b# #0.25 + 2=b# #2.25=b# Equation of the tangent line ... #y=-0.25x+2.25# To get a visual with the calculator solve the original equation for #y#. #y=(9-x^3)^(1/3)#

Adam Adkins · Answer

undefined To use implicit differentiation to find the equation of the tangent line to the curve $x^3 + y^3 = 9$ at the point where $x = -1$:

1. Differentiate both sides of the equation with respect to $x$.
2. Solve for $\frac{dy}{dx}$.
3. Plug in the given value $x = -1$ to find the slope of the tangent line.
4. Use the point-slope form of the equation of a line $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the given point, to find the equation of the tangent line.