Q.

ddxlog⁡exx−2x+23/4 is 

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a

x2+1x2−4

b

1

c

exx2−1x2−4

d

x2−1x2−4

answer is C.

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Detailed Solution

let y=log⁡exx−2x+23/4
=log⁡ex+log⁡x−2x+23/4⇒y=x+34[log⁡(x−2)−log⁡(x+2)]
On differentiating w.r.t. x, we get
dydx=ddxx+34{log⁡(x−2)−log⁡(x+2)}=1+341x−2−1x+2=1+3x2−4⇒dydx=x2−1x2−4

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