A(y+c) =b(y-c) solve for y
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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.
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