How do you find the derivative of #f(t)=t^(2/3)-t^(1/3)+4#?
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To find the derivative of ( f(t) = t^{2/3} - t^{1/3} + 4 ), you can use the power rule for derivatives. The power rule states that if ( f(t) = t^n ), then ( f'(t) = nt^{n-1} ). Applying this rule to each term in the function ( f(t) ), you get:
[ f'(t) = \frac{2}{3}t^{\frac{2}{3} - 1} - \frac{1}{3}t^{\frac{1}{3} - 1} + 0 ]
Simplify the exponents and constants:
[ f'(t) = \frac{2}{3}t^{-\frac{1}{3}} - \frac{1}{3}t^{-\frac{2}{3}} ]
Recall that ( t^{-n} = \frac{1}{t^n} ), so:
[ f'(t) = \frac{2}{3} \frac{1}{t^{\frac{1}{3}}} - \frac{1}{3} \frac{1}{t^{\frac{2}{3}}} ]
[ f'(t) = \frac{2}{3} \frac{1}{\sqrt[3]{t}} - \frac{1}{3} \frac{1}{\sqrt[3]{t^2}} ]
[ f'(t) = \frac{2}{3\sqrt[3]{t}} - \frac{1}{3\sqrt[3]{t^2}} ]
So, the derivative of ( f(t) ) is ( f'(t) = \frac{2}{3\sqrt[3]{t}} - \frac{1}{3\sqrt[3]{t^2}} ).
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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.
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.
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.
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.
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