Answer
$y'=\dfrac{3x^{2}(\sec^{2}\sqrt{1+x^{3}})\cos(\tan\sqrt{1+x^{3}})}{2\sqrt{1+x^{3}}}$
Work Step by Step
$y=\sin(\tan\sqrt{1+x^{3}})$
Start the differentiation process by using the chain rule:
$y'=(\tan\sqrt{1+x^{3}})'\cos(\tan\sqrt{1+x^{3}})=...$
Use the chain rule one more time to evaluate the indicated derivative:
$...=(\sqrt{1+x^{3}})'(\sec^{2}\sqrt{1+x^{3}})\cos(\tan\sqrt{1+x^{3}})=...$
Rewrite $\sqrt{1+x^{3}}$ using a rational exponent instead of the square root:
$...=[(1+x^{3})^{1/2}]'(\sec^{2}\sqrt{1+x^{3}})\cos(\tan\sqrt{1+x^{3}})=...$
Apply the chain rule one more time to evaluate the indicated derivative:
$...=\dfrac{1}{2}(1+x^{3})^{-1/2}(1+x^{3})'(\sec^{2}\sqrt{1+x^{3}})\cos(\tan\sqrt{1+x^{3}})=...$
$...=\dfrac{1}{2}(1+x^{3})^{-1/2}(3x^{2})(\sec^{2}\sqrt{1+x^{3}})\cos(\tan\sqrt{1+x^{3}})=...$
Change the sign of the negative exponent by moving its corresponding factor to the denominator:
$...=\dfrac{3x^{2}(\sec^{2}\sqrt{1+x^{3}})\cos(\tan\sqrt{1+x^{3}})}{2(1+x^{3})^{1/2}}=...$
$...=\dfrac{3x^{2}(\sec^{2}\sqrt{1+x^{3}})\cos(\tan\sqrt{1+x^{3}})}{2\sqrt{1+x^{3}}}$
