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  • 5 years ago

what is the derivative of 5cos^2(pi)(t)?

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  1. anonymous
    • 5 years ago
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    i assume you mean 5cos^2((pi*t)) in which case; i would use the chain rule, subs u=cos(pi*t), so you have (i presume it's y), y=5u^2 dy/du = 10u and du/dt=-pi*sin(pi*t) Multiply them together, dy/dx = 10u * -pi*sin(pi*t) replace u=cos(pi*t) dy/dx=-10*pi*cos(pi*t)*sin(pi*t) And you could then use the relation that sin(2x)=2sin(x)cos(x) so dy/dx=-5*pi*sin(2*pi*t) You could alternatively use the relation that 2cos(2x)=cos^2(x)+1 and rearrange for cos^2(x) [x being pi*t] .. which is a better way if you know how to do it .. hope this helps :)

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