## Calculus: Early Transcendentals 8th Edition

$J'(v)=1+\dfrac{1}{v^{2}}+\dfrac{6}{v^{4}}$
$J(v)=(v^{3}-2v)(v^{-4}+v^{-2})$ Differentiate using the product rule $J'(v)=(v^{3}-2v)(v^{-4}+v^{-2})'+(v^{-4}+v^{-2})(v^{3}-2v)'=...$ Evaluate the products and simplify $...=(v^{3}-2v)(-4v^{-5}-2v^{-3})+(v^{-4}+v^{-2})(3v^{2}-2)=...$ $...=-4v^{-2}-2+8v^{-4}+4v^{-2}+3v^{-2}-2v^{-4}+3-2v^{-2}=...$ $...=1+v^{-2}+6v^{-4}=1+\dfrac{1}{v^{2}}+\dfrac{6}{v^{4}}$