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The function is $\"f(t)=\\sqrt[3]{1+\\tan$ .

Consider $\"h(t)=\\sqrt[3]{1+t}\"$ and $\"g(t)=\\tan$ .

The function is in the form of composite function $\"h\\$ .

Consider $\"h(t)=\\sqrt[3]{1+t}\"$ .

Differentiate on each side with respect to $\"t\"$ .

$\"\\\\\\frac{d}{dt}h(t)=\\frac{d}{dt}(1+t)^{\\frac{1}{3}}\\\\h\'(t)=\\frac{1}{3}(1+t)^{\\frac{1}{3}-1}\\\\$

Consider $\"g(t)=\\tan$ .

Differentiate on each side with respect to $\"t\"$ .

$\"\\frac{d}{dt}g(t)=\\frac{d}{dt}\\tan$

$\"g\'(t)=\\sec^2$

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$\"f(t)=\\sqrt[3]{1+\\tan$

Chain Rule of derivatives: $\"\\frac{d}{dt}h(g(t))=$ .

Substitute $\"\\\\h\'(t)=\\frac{1}{3}(1+t)^{-\\frac{2}{3}}\"$ and $\"g\'(t)=\\sec^2$ in chain rule.

$\"\\\\f\'(t)=\\frac{1}{3}(1+\\tan$

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$\"\\\\f\'(t)=\\frac{1}{3\\sqrt[3]{(1+\\tan$ .