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Question

Find f (t){\vec{f}} \ '(t) for the vector function, and state the domain of each derived function.

f(t)=32t2iˉ+2t32t+4j^\vec{f}(t)=3^{2 t^2} \bar{i}+\dfrac{2 t-3}{2 t+4} \hat{j}

Solution

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Given the vector function

f(t)=32t2i^+2t32t+4j^.f(t)=3^{2t^2}\hat{i}+\dfrac{2t-3}{2t+4}\hat{j}.

Then

f(t)=2t232t214ti^+2(2t+4)2(2t3)(2t+4)2j^=8t332t21i^+14(2t+4)2j^\begin{aligned} f'(t) &=& 2t^23^{2t^2-1}4t\hat{i} +\dfrac{2(2t+4)-2(2t-3)}{(2t+4)^2}\hat{j}\\ &=& 8t^33^{2t^2-1}\hat{i}+\dfrac{14}{(2t+4)^2}\hat{j} \end{aligned}

where the first equality holds from the Chain rule and Leibniz rule and the second one comes from the division rule.

The domain of the derived function f(t)f'(t) is R{2}\mathbb{R}\setminus\{-2\}.

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