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fra1234

  • 3 years ago

Can you find dy/dx of y=log(sin^2x) step by step please?

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  1. Clifmak
    • 3 years ago
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    feeling a little bit to lazy to work it out,but can tell you the procedure.first find the derivative of sin^2x then use the general method for finding the derivative of a logarithmic function.

  2. adamshai
    • 3 years ago
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    \[\frac{ d }{ dx } {\log (\sin^2(2x))} = \frac{ 1 }{ \sin^2(2x) } \frac{ d }{ dx } \sin^2(2x) \\ = \frac{ 1 }{ \sin^2(2x) } \cdot 2\sin (2x) \cdot \frac{ d }{ dx } \sin (2x) \\ = \frac{ 1 }{ \sin^2(2x) } \cdot 2\sin (2x) \cdot \cos (2x) \frac{ d }{ dx } 2x \\ = \frac{ 1 }{ \sin^2(2x) } \cdot 2\sin (2x) \cdot \cos (2x) \cdot 2 \\ = \frac{ 4\cos (2x) }{ \sin(2x) } \\ = 4 \tan^{-1}(2x)\]

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