## Calculus: Early Transcendentals 8th Edition

$F'(t)=\dfrac{t^{4}+4t}{2(t^{3}+1)^{3/2}}$
$F(t)=\dfrac{t^{2}}{\sqrt{t^{3}+1}}$ Let's rewrite the function like this: $F(t)=\dfrac{t^{2}}{(t^{3}+1)^{1/2}}$ Differentiate using the quotient rule: $F'(t)=\dfrac{[(t^{3}+1)^{1/2}](t^{2})'-(t^{2})[(t^{3}+1)^{1/2}]'}{[(t^{3}+1)^{1/2}]^{2}}=...$ Use the chain rule to find $[(t^{3}+1)^{1/2}]'$ $...=\dfrac{[(t^{3}+1)^{1/2}](2t)-(t^{2})[\dfrac{1}{2}(t^{3}+1)^{-1/2}(t^{3}+1)']}{[(t^{3}+1)^{1/2}]^{2}}=...$ $...=\dfrac{2t(t^{3}+1)^{1/2}-(t^{2})[\dfrac{1}{2}(t^{3}+1)^{-1/2}(3t^{2})]}{[(t^{3}+1)^{1/2}]^{2}}=...$ Simplify: $...=\dfrac{2t(t^{3}+1)^{1/2}-\dfrac{3}{2}t^{4}(t^{3}+1)^{-1/2}}{t^{3}+1}=...$ $...=\dfrac{2t(t^{3}+1)^{1/2}-\dfrac{3t^{4}}{2(t^{3}+1)^{1/2}}}{t^{3}+1}=\dfrac{\dfrac{4t(t^{3}+1)-3t^{4}}{2(t^{3}+1)^{1/2}}}{t^{3}+1}=...$ $...=\dfrac{4t^{4}+4t-3t^{4}}{2(t^{3}+1)^{1/2}(t^{3}+1)}=\dfrac{t^{4}+4t}{2(t^{3}+1)^{3/2}}$