What is the derivative of #f(t) = (te^t +e^t, e^t-cost ) #?

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Answer 1 · 2017-05-07T19:27:52

#f'(t)=(e^t(t+2),e^t+sint)#

Take the derivative of the #x# component and #y# component separately.

That is, if #f(t)=(x(t),y(t))# then #f'(t)=(x'(t),y'(t))#.

So:

#x(t)=te^t+e^t=e^t(t+1)#

#x'(t)=e^t(t+1)+e^t(1)=e^t(t+2)#

And

#y(t)=e^t-cost#

#y'(t)=e^t+sint#

So:

#f'(t)=(e^t(t+2),e^t+sint)#