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Mathematics 17 Online
OpenStudy (anonymous):

find the derivative of y=sin^(2)(x)cos^(2)(x)

OpenStudy (anonymous):

save time and find the derivative of \(\frac{1}{4}\sin^2(2x)\)

OpenStudy (anonymous):

you got this?

OpenStudy (anonymous):

Y=[(SIN 2X)/2]^2 Y'= SIN2X(COS2X)(2)

sam (.sam.):

\[\text{Use the product rule, }\frac{d}{dx}(u v)=v \frac{du}{dx}+u \frac{dv}{dx}\text{, where }u=\cos ^2(x)\text{ and }v=\sin ^2(x)\]

OpenStudy (anonymous):

you will need to apply chain rule on sin^(2)(x) and cos^(2)(x) then apply product rule

OpenStudy (anonymous):

so for sin^(2)(x) g(x) = x^(2) g'(x) = 2x d(x) = sin(x) d'(x) = cos(x) thus the derivative of sin^(2)(x) is 2sin(x)cos(x) for cos^(2)(x) s(x) = x^(2) s'(x) = 2x l(x) = cos(x) l'(x) = -sin(x) thus the derivative of cos^(2)(x) = -2cos(x)sin(x) Now we apply product rule f'(x) = 2sin(x)cos(x)cos^(2)(x) + (-2cos(x)sin(x)sin^(2)(x))

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