Product Rule Shortcut for Complicated Derivatives

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SUMMARY

The discussion focuses on finding the derivative of the function y = (x² + 1)⁷(x⁹ + 2)⁵(x³ + 1)³(x⁸ + 7)³ using shortcuts to simplify the process. Participants suggest that while the product rule is necessary, utilizing the chain rule in conjunction with logarithmic differentiation can streamline the calculations. By taking the natural logarithm of both sides, the derivative can be expressed as a sum of simpler terms, which can then be multiplied back by y to find y'. This method reduces the complexity of applying the product rule multiple times.

PREREQUISITES
  • Understanding of the Product Rule in calculus
  • Familiarity with the Chain Rule in calculus
  • Knowledge of logarithmic differentiation techniques
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study advanced applications of the Product Rule in calculus
  • Learn about logarithmic differentiation in detail
  • Explore the Chain Rule and its applications in complex functions
  • Practice solving derivatives of polynomial products using both product and chain rules
USEFUL FOR

Students and educators in calculus, mathematicians tackling complex derivatives, and anyone looking to enhance their differentiation techniques for polynomial functions.

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Find y'
y=(x2+1)7(x9+2)5(x3+1)3(x8+7)3

Is there a shortcut to doing this problem? Or do I have to actually use the product rule more than 3 times?
 
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The easiest way would be to just use the product rule. Using the chain rule might make it a bit easier in terms of making sure that you don't make a mistake.
 
You could take logs of each side:
<br /> \begin{align*}<br /> \ln y &amp; = \ln{\left(x^2+1\right)^7 \left(x^9+2\right)^5 \left(x^3+1\right)^3 \left(x^8 + 7\right)^3} \\<br /> &amp; = 7\ln(x^2+1) + 5\ln(x^9+1) + 3\ln(x^3+1) + 3\ln(x^8+7)<br /> \end{align*}<br />

Differentiating gives
<br /> \frac{y&#039;}{y}<br />

on the left and a sum of terms on the right: multiply through by y and cancelling terms may make the work slightly more palatable.
 

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