## Calculus: Early Transcendentals 8th Edition

$y'=\dfrac{4+\ln(t^{4})}{2\sqrt{t\ln(t^{4})}}$
$y=\sqrt{t\ln(t^{4})}$ Rewrite the square root using a rational exponent: $y=[t\ln(t^{4})]^{1/2}$ Start the differentiation process by using the chain rule: $y'=\dfrac{1}{2}[t\ln(t^{4})]^{-1/2}[t\ln(t^{4})]'=...$ Change the sign of the negative exponent by changing its corresponding factor to the denominator: $...=\dfrac{[t\ln(t^{4})]'}{2[t\ln(t^{4})]^{1/2}}=...$ Rewrite the rational exponent in the denominator as a square root: $...=\dfrac{[t\ln(t^{4})]'}{2\sqrt{t\ln(t^{4})}}$ Use the product rule to evaluate the derivative indicated in the numerator: $...=\dfrac{t[\ln(t^{4})]'+(t)'\ln(t^{4})}{2\sqrt{t\ln(t^{4})}}=...$ Use the chain rule one last time to evaluate the derivatives indicated in the numerator and simplify: $...=\dfrac{t\Big[\dfrac{(t^{4})'}{t^{4}}\Big]+\ln(t^{4})}{2\sqrt{t\ln(t^{4})}}=\dfrac{t\Big(\dfrac{4t^{3}}{t^{4}}\Big)+\ln(t^{4})}{2\sqrt{t\ln(t^{4})}}=\dfrac{4+\ln(t^{4})}{2\sqrt{t\ln(t^{4})}}$