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

Answer 1

Isaiah Allen

#dy/dx=(t(2t+sint))/(2t^2-1)#

Let's call #x(t)=t^2-lnt#

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

and #y(t)=t^2-cos^2t#

Then #y'(t)=dy/dt=2t-2cost*-sint=2t+sin2t#

And finally #dy/dx=(dy/dt)/(dx/dt)=(2t +sint)/((2t^2-1)/t)#

#dy/dx=(t(2t+sint))/(2t^2-1)#

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Answer 2

Cameron Alexander

To find the derivative of the vector-valued function ( \mathbf{f}(t) = (t^2 - \ln t, t^2 - \cos^2 t) ), we differentiate each component of the vector function separately with respect to ( t ).

The derivative of the first component ( t^2 - \ln t ) with respect to ( t ) is ( 2t - \frac{1}{t} ).

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

Therefore, the derivative of ( \mathbf{f}(t) ) is ( \mathbf{f}'(t) = (2t - \frac{1}{t}, 2t + 2\cos t \sin 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.