Solution assignment 27 Chain rule

Assignment 27

Differentiate:

$y=x\ln(\displaystyle\frac{1}{x})$

Solution

If we do not simplify the function we have to use the product rule in combination with the informal chain rule (the 'quote' means: the derivative of):

$y'=1\cdot\ln(\displaystyle\frac{1}{x})+x[\ln(\displaystyle\frac{1}{x})]'$

We can write:

$[\ln(\displaystyle\frac{1}{x})]'=\displaystyle\frac{1}{\displaystyle\frac{1}{x}}\cdot(\displaystyle\frac{1}{x})'=x\cdot(x^{-1})'=x\cdot-x^{-2}=-\displaystyle\frac{1}{x}$

and thus we get:

$y'=\ln(\displaystyle\frac{1}{x})-1$

Note that we could have written:

$[\ln(\displaystyle\frac{1}{x}]'=[\ln(x^{-1})]'=-\displaystyle\frac{1}{x}$

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Solution assignment 27 Chain rule

Assignment 27

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