How do you differentiate F(y)=(1/y^2-3/y^4)(y+5y^3)?

Jul 19, 2017

$\frac{d}{\mathrm{dy}} F \left(y\right) = 15 {y}^{2} + \frac{59}{y} ^ 2 + \frac{9}{y} ^ 4 - 55$

Explanation:

$F \left(y\right) = \left(\frac{1}{y} ^ 2 - \frac{3}{y} ^ 4\right) \left(y + 5 {y}^{3}\right)$

Since, $\frac{d}{\mathrm{dx}} g \left(x\right) f \left(x\right) = g \left(x\right) f ' \left(x\right) + f \left(x\right) g ' \left(x\right)$

$\frac{d}{\mathrm{dy}} F \left(y\right) = \left(\frac{1}{y} ^ 2 - \frac{3}{y} ^ 4\right) \frac{d}{\mathrm{dy}} \left(y + 5 {y}^{3}\right) + \frac{d}{\mathrm{dy}} \left(\frac{1}{y} ^ 2 - \frac{3}{y} ^ 4\right) \left(y + 5 {y}^{3}\right)$

$\frac{d}{\mathrm{dy}} F \left(y\right) = \left(\frac{1}{y} ^ 2 - \frac{3}{y} ^ 4\right) \left(1 + 15 {y}^{4}\right) + \left(- \frac{2}{y} ^ 3 + \frac{12}{y} ^ 5\right) \left(y + 5 {y}^{3}\right)$

$\frac{d}{\mathrm{dy}} F \left(y\right) = \left(\frac{1}{y} ^ 2 - \frac{3}{y} ^ 4 - 45 + 15 {y}^{2}\right) + \left(- \frac{2}{y} ^ 2 + \frac{12}{y} ^ 4 - 10 + \frac{60}{y} ^ 2\right)$

$\frac{d}{\mathrm{dy}} F \left(y\right) = \frac{1}{y} ^ 2 - \frac{3}{y} ^ 4 - 45 + 15 {y}^{2} - \frac{2}{y} ^ 2 + \frac{12}{y} ^ 4 - 10 + \frac{60}{y} ^ 2$

$\frac{d}{\mathrm{dy}} F \left(y\right) = 15 {y}^{2} + \frac{59}{y} ^ 2 + \frac{9}{y} ^ 4 - 55$