Difference between revisions of "022 Exam 2 Sample B, Problem 3"
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!Step 1:  !Step 1:  
    
−  We need to start by identifying the two functions that are being multiplied together so we can apply the product rule.  +  We need to start by identifying the two functions that are being multiplied together so we can apply the product rule. Let's call <math style="verticalalign: 20%">g(x)\,=\,2x^3,\,</math> and <math style="verticalalign: 20%">\,h(x) \, = \, e^{3x + 5}</math>, so <math style="verticalalign: 20%">f(x)=g(x)\cdot h(x)</math>. 
−  
−  
−  :  
}  }  
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!Step 2:  !Step 2:  
    
−  We can now apply the  +  We can now apply the advanced techniques.This allows us to see that 
    
    
−  <math>\begin{array}{rcl}  +  ::<math>\begin{array}{rcl} 
f'(x)&=&2(x^3)' e^{3x+5}+2x^3(e^{3x+5})' \\  f'(x)&=&2(x^3)' e^{3x+5}+2x^3(e^{3x+5})' \\  
&=&6x^2e^{3x+5}+2x^3(3e^{3x+5})\\  &=&6x^2e^{3x+5}+2x^3(3e^{3x+5})\\  
−  & = &6x^2e^{3x+5}+6x^3e^{3x+5}  +  & = &6x^2e^{3x+5}+6x^3e^{3x+5}. 
\end{array}</math>  \end{array}</math>  
}  }  
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−  <math>6x^2e^{3x+5}+6x^3e^{3x+5}  +  ::<math>f'(x)\,=\,6x^2e^{3x+5}+6x^3e^{3x+5}. 
</math>  </math>  
}  }  
[[022_Exam_2_Sample_B'''<u>Return to Sample Exam</u>''']]  [[022_Exam_2_Sample_B'''<u>Return to Sample Exam</u>''']] 
Latest revision as of 07:50, 17 May 2015
Find the derivative of .
Foundations:  

This problem requires several advanced rules of differentiation. In particular, you need  
The Chain Rule: If and are differentiable functions, then  
The Product Rule: If and are differentiable functions, then  
Additionally, we will need our power rule for differentiation:  
 
as well as the derivative of the exponential function, :  

Solution:
Step 1: 

We need to start by identifying the two functions that are being multiplied together so we can apply the product rule. Let's call and , so . 
Step 2: 

We can now apply the advanced techniques.This allows us to see that 

Final Answer: 

