Find the derivative of function y=`log_(10)` `(x^2-1)/(x)`



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Posted on (Answer #1)

`y=log_10 (x^2-1)/x`

Expand right side. Apply the property of logarithm which is `log_b M/N=log_b M - log_b N` .

`y=log_10 (x^2-1) - log_10 x`

To determine y', take the derivative of each term.

`y'=(log_10 (x^2-1))' - (log_10x)'`

Apply the formula of derivative of logarithmic functions which is `(log_b u)' = (log_b e)/u*u'` .

`y'=(log_10 e)/(x^2-1)*(x^2-1)' - (log_10 e)/x*x'`

` y'=(log_10 e)/(x^2-1)*2x - (log_10 e)/x*1 `

`y'=(2xlog_10 e)/(x^2-1)-(log_10 e)/x`

Hence, the derivative of the given function is `y'=(2xlog_10 e)/(x^2-1)-(log_10 e)/x` .

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