Is the derivative of $\displaystyle g(t)=t^2+sint+cos\frac{\pi}{4}$ $\displaystyle g'(t)=2t+cost$ ?
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Originally Posted by VonNemo19 Is the derivative of $\displaystyle g(t)=t^2+sint+cos\frac{\pi}{4}$ $\displaystyle g'(t)=2t+cost$ ? Yes it is!
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