Question

the derivative of sinx w.r.t log x is​

Answer 1

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Answer 2

Answer:

The derivative is $\frac{xlogxcosx-sinx\ }{2xlogx} }$.

Step-by-step explanation:

We need to find the derivative of $sinx$ w.r.t $log x$ is​

Mathematically,

$\frac{sinx}{logx} dx$

We will use quotient rule here:

$\frac{u}{v} dx=\frac{vu^{'}-uv^{'} }{v^{2} }$

Applying it here, we get

$=\frac{logxsinx^{'}-sinxlogx^{'} }{logx^{2} }$

$=\frac{logxcosx-sinx\frac{1}{x} }{logx^{2} }$

$=\frac{xlogxcosx-sinx\ }{xlogx^{2} }$

We know the property of log that

$logm^{n} =nlogm$

$=\frac{xlogxcosx-sinx\ }{2xlogx} }$

Hence, the final value is $\frac{xlogxcosx-sinx\ }{2xlogx} }$.

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