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

Question

Isabella Ackerman · Accepted Answer

#8/5# #d/(dx)(x^3+y^3=4xy+1)--(1)# we will differentiate #wrt" "x#. Remembering that when differentiating #y# we multiply by #(dy)/(dx)# by virtue of the chain rule. Also on the #RHS # we will need the product rule on the first term #(1)rarr3x^2+3y^2(dy)/(dx)=4y+4x(dy)/(dx)+0# rearrange for #(dy)/(dx)# #3y^2(dy)/(dx)-4x(dy)/(dx)=4y-3x^2# #(dy)/(dx)(3y^2-4x)=4y-3x^2# #=>(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#

Charles Arocha · Answer

undefined To find $ \frac{{dy}}{{dx}} $ by implicit differentiation of $ x^3 + y^3 = 4xy + 1 $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Apply the chain rule when differentiating terms involving $ y $.
3. Solve for $ \frac{{dy}}{{dx}} $.
4. Evaluate $ \frac{{dy}}{{dx}} $ at the given point (2,1).

Differentiating both sides of the equation $ x^3 + y^3 = 4xy + 1 $ with respect to $ x $, we get:

$$ 3x^2 + 3y^2 \frac{{dy}}{{dx}} = 4x + 4y \frac{{dy}}{{dx}} $$

Rearrange the terms to isolate $ \frac{{dy}}{{dx}} $:

$$ 3y^2 \frac{{dy}}{{dx}} - 4y \frac{{dy}}{{dx}} = 4x - 3x^2 $$

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

$$ \frac{{dy}}{{dx}} = \frac{{4x - 3x^2}}{{3y^2 - 4y}} $$

Now, evaluate $ \frac{{dy}}{{dx}} $ at the point (2,1):

$$ \frac{{dy}}{{dx}} = \frac{{4(2) - 3(2)^2}}{{3(1)^2 - 4(1)}} $$

$$ \frac{{dy}}{{dx}} = \frac{{8 - 12}}{{3 - 4}} $$

$$ \frac{{dy}}{{dx}} = \frac{{-4}}{{-1}} $$

$$ \frac{{dy}}{{dx}} = 4 $$

So, $ \frac{{dy}}{{dx}} = 4 $ at the point (2,1).