Solution assignment 27 Chain rule

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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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