Difference between revisions of "009A Sample Final 1, Problem 7"

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!Step 1:    
 
!Step 1:    
 
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|Using implicit differentiation on the equation <math>x^3+y^3=6xy</math>, we get
 
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|<math>3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}</math>.
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Now, we move all the <math>\frac{dy}{dx}</math> terms to one side of the equation.
 +
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|So, we have
 
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|<math>3x^2-6y=\frac{dy}{dx}(6x-3y^2)</math>.
 
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|We solve for <math>\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}</math>.
 
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Revision as of 16:53, 14 February 2016

A curve is defined implicityly by the equation

a) Using implicit differentiation, compute .

b) Find an equation of the tangent line to the curve at the point .

Foundations:  

Solution:

(a)

Step 1:  
Using implicit differentiation on the equation , we get
.
Step 2:  
Now, we move all the terms to one side of the equation.
So, we have
.
We solve for .

(b)

Step 1:  
Step 2:  
Step 3:  
Final Answer:  
(a)
(b)

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