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

Answer 1

Emily Ammerman

# f'(t) = (2t^2cost - t^3sint) / (2lnt - t ) #

Define #x(t)# and #y(t)# such that

# {(x(t) = ln^2t-t), (y(t) = t^2cost) :} #

Using the chain rule and product rule we can differentiate #x# and #y# wrt #t# as follows:

# dx/dt = 2(lnt)1/t - 1 #

# :. dx/dt = 2/tlnt - 1 #

And,

# dy/dt = (t^2)(-sint) + (2t)(cost) #

# :. dy/dt = 2tcost - t^2sint #

And Also, By the chain rule:

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

# :. dy/dx = (2tcost - t^2sint) / (2/tlnt - 1 ) #

# :. dy/dx = (2tcost - t^2sint) / (1/t(2lnt - t )) #

# :. dy/dx = (2t^2cost - t^3sint) / (2lnt - t ) #

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Answer 2

Charles Anctil

The derivative of ( f(t) = (\ln(t)^2 - t, t^2 \cos(t)) ) with respect to ( t ) is:

( f'(t) = \left(2\frac{\ln(t)}{t} - 1, 2t\cos(t) - t^2\sin(t)\right) )

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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.