differentiate. f(y) = (1/y^2 - 5/y^4)(y + 9y^3)

Explanation:

Step1: Apply product - rule

The product - rule states that if $F(y)=u(y)v(y)$, then $F^\prime(y)=u^\prime(y)v(y)+u(y)v^\prime(y)$. Let $u(y)=\frac{1}{y^{2}}-\frac{5}{y^{4}}=y^{-2}-5y^{-4}$ and $v(y)=y + 9y^{3}$.

Step2: Differentiate $u(y)$

Using the power - rule $\frac{d}{dy}(y^{n})=ny^{n - 1}$, we have $u^\prime(y)=-2y^{-3}+20y^{-5}=-\frac{2}{y^{3}}+\frac{20}{y^{5}}$.

Step3: Differentiate $v(y)$

Using the power - rule, $v^\prime(y)=1 + 27y^{2}$.

Step4: Apply the product - rule formula

$F^\prime(y)=u^\prime(y)v(y)+u(y)v^\prime(y)$ $=\left(-\frac{2}{y^{3}}+\frac{20}{y^{5}}\right)(y + 9y^{3})+\left(\frac{1}{y^{2}}-\frac{5}{y^{4}}\right)(1 + 27y^{2})$ $=\left(-\frac{2}{y^{3}}\times y-\frac{2}{y^{3}}\times9y^{3}+\frac{20}{y^{5}}\times y+\frac{20}{y^{5}}\times9y^{3}\right)+\left(\frac{1}{y^{2}}\times1+\frac{1}{y^{2}}\times27y^{2}-\frac{5}{y^{4}}\times1-\frac{5}{y^{4}}\times27y^{2}\right)$ $=\left(-\frac{2}{y^{2}}-18+\frac{20}{y^{4}} + 180y^{-2}\right)+\left(\frac{1}{y^{2}}+27-\frac{5}{y^{4}}-\frac{135}{y^{2}}\right)$ $=\left(-\frac{2}{y^{2}}+\frac{180}{y^{2}}+\frac{1}{y^{2}}-\frac{135}{y^{2}}\right)+\left(\frac{20}{y^{4}}-\frac{5}{y^{4}}\right)+(-18 + 27)$ $=\frac{-2 + 180+1-135}{y^{2}}+\frac{20 - 5}{y^{4}}+9$ $=\frac{44}{y^{2}}+\frac{15}{y^{4}}+9$

Answer:

$F^\prime(y)=9+\frac{44}{y^{2}}+\frac{15}{y^{4}}$