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What is the derivative of #g(t)=(pi)(cos t - 1/t^2)#?

1 Answer

Aug 31, 2017

#=pi(2/(t^3) - sin(t))#

Explanation:

remember that #pi# is a constant, so you can just mutiply through:

#g(t) = pi*cos(t) - pi/(t^2) = pi * cos(t) - pi * t^-2#

#(dg)/dt = -pi sin(t) + 2pit^-3#

#=pi(2/(t^3) - sin(t))#

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