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Question

Find the derivative of exsinx.

A
excosec x[cotx+1]
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B
excosec x[1cotx]
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C
exsecx[cotx+1]
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D
exsecx[cotx1]
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Solution

The correct option is C excosec x[1cotx]
On Differentiation we get
ddxexcosecx=ex.cosecxcotx+ex.cosecx=ex.cosecx[1cotx]

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