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B
2ln√2
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C
−√2ln√2
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D
√2ln√2
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Solution
The correct option is A−2ln√2 x=π4⇒y=0
We can write, y=logcosx(tanx)⇒y=logcosxsinxcosx⇒y=logcosxsinx−1⇒y+1=lnsinxlncosx⇒(y+1)(lncosx)=lnsinx
Differentiating w.r.t. x, ⇒(y+1)×−sinxcosx+y′(lncosx)=cosxsinx
Putting x=π4,y=0 ⇒−1−y′(ln√2)=−1⇒y′=−2ln√2