How do you find the derivative of 1/logx?

Mathematics

1 answer:

valentina_108 [34]11 months ago

5 0

The derivative of 1/logx is With the chain rule.

1log(x)=log(x)−1 is ,= -1xlog(x)2 .

The by-product of logₐ x (log x with base a) is 1/(x ln a). Here, the thrilling issue is that we have "ln" withinside the by-product of "log x". Note that "ln" is referred to as the logarithm (or) it's miles a logarithm with base "e".

The by-product of 1/log x is -1/x(log x)^2. Note that 1/logx is the reciprocal of log.

$With the chain rule. \\ 1log(x)=log(x)−1 \\ So you follow the chain rule to the electricity after which to the log: \\ ddxlog(x)−1=−1⋅log(x) \\ \\−2⋅ddxlog(x) \\ =−1log(x)21x \\ =−1xlog(x)2 .$

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