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Differentiate logsinx with respect to x ...

Differentiate `logsinx` with respect to `x` :

Text Solution

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We have given: `logsinx`
Let `y=logsinx`
Now differentiating w.r.to `x`we get
`(dy)/(dx)=d/(dx)(logsinx)`
`implies(dy)/(dx)=1/(sinx).d/(dx)(sinx)`
`implies(dy)/(dx)=1/(sinx).cosx`
`implies(dy)/(dx)=cotx`
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