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Question from a Calc 2 student.

To solve the following derivative $$\frac{d}{dn}\left(\frac{1}{2^{2n+1}}\frac{\left(\frac{1}{2}\right)^{2n+2}}{\left(2n+1\right)\left(2n+2\right)}\right)$$

can a dummy variable be used for $2n+2$ or $2n+1$ to make this more manageable?

kimchi lover
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findingmyway
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  • Yes you can use those dummy variables. More importantly though you can and should also simplify the fraction by moving the exponent above tho line into the denominator - this will make this derivative look less terrific. – Guenterino Dec 12 '20 at 19:09

1 Answers1

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If you let, say, u= 2n+ 1 and v= 2n+ 2 then your function becomes $f(u,v)= \frac{1}{2^u}\frac{\left(\frac{1}{2}\right)^v}{uv}= \frac{2^{-u}2^{-v}}{uv}$.

Now you need to use the "chain rule": $\frac{df}{dn}= \frac{\partial f}{\partial u}\frac{du}{dn}+ \frac{\partial f}{\partial v}\frac{dv}{dn}$.

(I am surprisingly uncomfortable using "n" as a continuous variable!)

user247327
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