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

Answer 1

Evelyn Agard

f'(t) = #((2 ln t)/t-1, csc t (1-t cot t))#

f(t) = ( x(t), y(t0). f'(t) = ( x'(t), y'(t)).

x(t) = #(ln t )^2-t = u^2 -t, u=ln t#.. Applying function of function rule, f'(t)=2uu'(t)-1= #2( ln t)/t-1#.

y(t) = t csc t. Applying product rule, y'(t) = #csct-t csc t cot t=csc t (1-t cot t)#.#

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Answer 2

Aurora Alvarado

The derivative of ( f(t) = (\ln(t))^2 - t , \text{csc}(t) ) with respect to ( t ) is:

[ f'(t) = 2\ln(t)\cdot\frac{1}{t} - 1 \cdot \text{csc}(t) - t \cdot (-\text{csc}(t)\cdot\cot(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.