What is the derivative of #f(t) = (t-lnt^2, t^2-sint ) #?

Answer 1

Charles Anctil

#dy/dx=(2t^2-tcost)/(t-2)#

Given #x=f(t),y=g(t)#

#dy/dx=(dy/dt)/(dx/dt)#

#dy/dt=g'(t)=2t-cost#

#dx/dt=f'(t)=1-(2t)/t^2=1-2/t#

#dy/dx=(2t-cost)/(1-2/t)=(2t-cost)/((t-2)/t)=(2t^2-tcost)/(t-2)#

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Answer 2

Charles Anctil

To find the derivative of f(t) = (t - ln(t^2), t^2 - sin(t)), you differentiate each component of the vector function separately with respect to t.

The derivative of the first component, t - ln(t^2), with respect to t is 1 - (2/t).

The derivative of the second component, t^2 - sin(t), with respect to t is 2t - cos(t).

So, the derivative of f(t) is the vector function:

f'(t) = (1 - (2/t), 2t - cos(t))

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.