Calculus: Early Transcendentals 8th Edition

$y'=\dfrac{3x^{2}(\sec^{2}\sqrt{1+x^{3}})\cos(\tan\sqrt{1+x^{3}})}{2\sqrt{1+x^{3}}}$
$y=\sin(\tan\sqrt{1+x^{3}})$ 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: $...=\dfrac{3x^{2}(\sec^{2}\sqrt{1+x^{3}})\cos(\tan\sqrt{1+x^{3}})}{2(1+x^{3})^{1/2}}=...$ $...=\dfrac{3x^{2}(\sec^{2}\sqrt{1+x^{3}})\cos(\tan\sqrt{1+x^{3}})}{2\sqrt{1+x^{3}}}$