How do you differentiate #4xy-3x-11=0#?

Question

Isaiah Allen · Accepted Answer

#dy/dx=(3-4y)/(4x)#

Assume the equation #4xy-3x-11=0# implicitly defines #y# as a function of #x# and then differentiate with respect to #x#, using the Product Rule:

#4y+4x dy/dx-3=0#

Now solve for #dy/dx# to get #dy/dx=(3-4y)/(4x)#.

If you happen to know a specific point #(x,y)# on the curve defined by the original equation #4xy-3x-11=0#, you can plug the coordinates of that point into #dy/dx=(3-4y)/(4x)# to find the slope of the curve at that point.

The particular equation #4xy-3x-11=0# is simple enough that we can check our work another way; we can solve for #y# explictly as a function of #x#:

#4xy-3x-11=0\Rightarrow y=(3x+11)/(4x)=3/4+11/4 x^{-1}#

Now we can differentiate this in the ordinary way to get #dy/dx=-11/4 x^{-2}=-11/(4x^2)#. Is this answer the same as the original? It sure looks different. It can be seen to be the same answer by substituting #y=(3x+11)/(4x)# into #dy/dx=(3-4y)/(4x)# in place of #y# and simplifying:

#dy/dx=(3-4((3x+11)/(4x)))/(4x)=(3x-(3x+11))/(4x^2)=-11/(4x^2)#

It is the same!

As an example of a point on this curve, we can use the equation #y=(3x+11)/(4x)# to find #y# when #x=1# to be #y=14/4=7/2#, meaning that the point #(x,y)=(1,7/2)# is on the graph of #4xy-3x-11=0#. The slope of the curve at that point is #dy/dx=(3-14)/4=-11/4=-2.75#.

The equation of the tangent line to the curve at that point is therefore #y=-11/4(x-1)+7/2=-11/4 x+25/4#.

Here's a graph of the situation just described:

Isaiah Allen · Answer

Assuming that we want to find #dy/dx#:

Using implicit differentiation, we get: #dy/dx = (3-4y)/(4x)#.

Solving for #y# first, we get: #dy/dx = 11/(4x^2)#.

The question is posted under "Implicit Differentiation", so let's do it that way first: #4xy-3x-11=0# Leaving the function Implicit In order to differentiate #4xy#, we will need the product rule. Remember that we are assuming that #y# is some function of #x#, so we have #4xy = 4xf(x)# and we use the product rule to get: the derivative is: #4f(x)+4xf'(x)# Back to this problem: #d/dx(4xy)-d/dx(3x)-d/dx(11)=d/dx(0)# #4y+4xdy/dx-3=0# #4xdy/dx = 3-4y# #dy/dx = (3-4y)/(4x)# Making the function explicit Solve #4xy-3x-11=0# for #y# #y = (3x+11)/(4x)# We could differentiate using the quotient rule, but it is perhaps simpler to rewrite again: #y = (3x)/(4x)+11/(4x)# # = 3/4 +11/4x^-1# So #dy/dx = -11/4x^-2# # = -11/(4x^2)# The answers are equivalent To see that the answer are equivalent compare: #dy/dx = (3-4y)/(4x)# with #y = (3x+11)/(4x)# and #dy/dx = -11/(4x^2)# Using Implicit differentiation, there is still a #y# in the derivative. That is the price we pay for not making the function explicit before differentiating. If we substitute the solution for #y#, we get: #dy/dx = (3-4((3x+11)/(4x)))/(4x)# # = (3-(3x+11)/x)/(4x)# # = (3x-3x-11)/(4x^2)# # = -11/(4x^2)#

Adam Adkins · Answer

undefined To differentiate $ 4xy - 3x - 11 = 0 $:

1. Rewrite the equation as $ 4xy - 3x = 11 $.
2. Differentiate both sides of the equation with respect to $ x $.
3. Apply the product rule to differentiate $ 4xy $ with respect to $ x $.
4. Differentiate $ -3x $ with respect to $ x $.
5. Solve for $ \frac{dy}{dx} $ by isolating $ y' $ on one side of the equation.

The result will be:

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