Answer
$H'(r)=\dfrac{2(r^{2}-1)^{2}(r^{2}+3r+5)}{(2r+1)^{6}}$
Work Step by Step
$H(r)=\dfrac{(r^{2}-1)^{3}}{(2r+1)^{5}}$
Differentiate using the quotient rule:
$H'(r)=\dfrac{[(2r+1)^{5}][(r^{2}-1)^{3}]'-[(r^{2}-1)^{3}][(2r+1)^{5}]'}{[(2r+1)^{5}]^{2}}=...$
Use the chain rule to find $[(r^{2}-1)^{3}]'$ and $[(2r+1)^{5}]'$:
$...=\dfrac{[(2r+1)^{5}][3(r^{2}-1)^{2}(r^{2}-1)']-[(r^{2}-1)^{3}][5(2r+1)^{4}(2r+1)']}{[(2r+1)^{5}]^{2}}=...$
$...=\dfrac{[(2r+1)^{5}][3(r^{2}-1)^{2}(2r)]-[(r^{2}-1)^{3}][5(2r+1)^{4}(2)]}{[(2r+1)^{5}]^{2}}=...$
Simplify:
$...=\dfrac{6r(2r+1)^{5}(r^{2}-1)^{2}-10(2r+1)^{4}(r^{2}-1)^{3}}{(2r+1)^{10}}=...$
Take out common factors $2$, $(2r+1)^{4}$ and $(r^{2}-1)^{2}$, then continue simplifying:
$...=\dfrac{2(2r+1)^{4}(r^{2}-1)^{2}[3r(2r+1)-5(r^{2}-1)]}{(2r+1)^{10}}=...$
$...=\dfrac{2(2r+1)^{4}(r^{2}-1)^{2}(6r^{2}+3r-5r^{2}+5)}{(2r+1)^{10}}=...$
$...=\dfrac{2(2r+1)^{4}(r^{2}-1)^{2}(r^{2}+3r+5)}{(2r+1)^{10}}=...$
$...=\dfrac{2(r^{2}-1)^{2}(r^{2}+3r+5)}{(2r+1)^{6}}$
