Product Rule Shortcut for Complicated Derivatives

tsaitea
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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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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