Difference between revisions of "022 Sample Final A, Problem 4"
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When we use implicit differentiation, we combine the chain rule with the fact that <math style="verticalalign: 18%">y</math> is a function of <math style="verticalalign: 0%">x</math>, and could really be written as <math style="verticalalign: 21%">y(x).</math> Because of this, the derivative of <math style="verticalalign:17%">y^3</math> with respect to <math style="verticalalign: 0%">x</math> requires the chain rule, so  When we use implicit differentiation, we combine the chain rule with the fact that <math style="verticalalign: 18%">y</math> is a function of <math style="verticalalign: 0%">x</math>, and could really be written as <math style="verticalalign: 21%">y(x).</math> Because of this, the derivative of <math style="verticalalign:17%">y^3</math> with respect to <math style="verticalalign: 0%">x</math> requires the chain rule, so  
    
−    +   
+  ::<math>\frac{d}{dx}\left(y^{3}\right)=3y^{2}\cdot\frac{dy}{dx}.</math>  
    
−  For this problem we also need to use the product rule.  +  For this problem, we also need to use the product rule. 
}  }  
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!Step 1:  !Step 1:  
    
−  First, we differentiate each term separately with respect to <math style="verticalalign:  +  First, we differentiate each term separately with respect to <math style="verticalalign: 0px">x</math> and apply the product rule on the right hand side to find that  <math style="verticalalign: 4px">x + y = x^3y^3</math>  differentiates implicitly to 
    
    
−  ::<math>1 + \frac{dy}{dx} = 3x^2y^3 + 3x^3y^2 \cdot \frac{dy}{dx}</math>.  +  ::<math>1 + \frac{dy}{dx} \,=\, 3x^2y^3 + 3x^3y^2 \cdot \frac{dy}{dx}</math>. 
}  }  
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!Step 2:  !Step 2:  
    
−  Now we need to solve for <math>\frac{dy}{dx}</math> and doing so we find that <math>\frac{dy}{dx} = \frac{3x^2y^3  1}{1  3x^3y^2}</math>  +  Now we need to solve for  <math style="verticalalign: 14px">\frac{dy}{dx}</math> , and doing so we find that <math style="verticalalign: 18px">\frac{dy}{dx} \,=\, \frac{3x^2y^3  1}{1  3x^3y^2}</math>. 
}  }  
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!Final Answer:  !Final Answer:  
    
−  <math>\frac{dy}{dx} = \frac{3x^2y^3  1}{1  3x^3y^2}</math>  +   
+  ::<math>\frac{dy}{dx} \,=\, \frac{3x^2y^3  1}{1  3x^3y^2}.</math>  
}  }  
[[022_Sample Final A'''<u>Return to Sample Exam</u>''']]  [[022_Sample Final A'''<u>Return to Sample Exam</u>''']] 
Latest revision as of 18:33, 6 June 2015
Use implicit differentiation to find
Foundations: 

When we use implicit differentiation, we combine the chain rule with the fact that is a function of , and could really be written as Because of this, the derivative of with respect to requires the chain rule, so 

For this problem, we also need to use the product rule. 
Solution:
Step 1: 

First, we differentiate each term separately with respect to and apply the product rule on the right hand side to find that differentiates implicitly to 

Step 2: 

Now we need to solve for , and doing so we find that . 
Final Answer: 

