question of differentiation
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srijandipdas:
please help me
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1
y = log tanx
You just have to apply the chain rule. See!
dy/dx = {d/dtanx (logtanx)}×{d/dx(tanx)}
= {1/ tanx}×{sec²x}
= {cosx/sinx cos²x}
= {1/sinx cosx}
= {2/2sinx cosx}
= 2/sin2x
= 2cosec2x
Hence the correct option is option 2.
You just have to apply the chain rule. See!
dy/dx = {d/dtanx (logtanx)}×{d/dx(tanx)}
= {1/ tanx}×{sec²x}
= {cosx/sinx cos²x}
= {1/sinx cosx}
= {2/2sinx cosx}
= 2/sin2x
= 2cosec2x
Hence the correct option is option 2.
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Answer:
2cosec2x
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