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Mathematics 18 Online
OpenStudy (anonymous):

Find derivative of sin^2(x)

OpenStudy (anonymous):

think of it as \[\left(\sin(x)\right)^2\] and use the chain rule

OpenStudy (anonymous):

How do i do that?

OpenStudy (anonymous):

the derivative of \(x^2\) is \(2x\) and by the chain rule the derivative of \(f^2(x)\) is \[2f(x)f'(x)\]

OpenStudy (anonymous):

put \(f(x)=\sin(x)\) and \(f'(x)=\cos(x)\) in the above

OpenStudy (anonymous):

Well like i knew that x^2=2x and sinx=cosx but i dont know how to work the problem and write the answer

OpenStudy (anonymous):

i am fairly sure that you don't believe either \(x^2=2x\) or \(\sin(x)=\cos(x)\)

OpenStudy (anonymous):

sen²x = sen x * sen x sen²x = 1 - cos²x is just a trigonometric identity

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