What is the derivative of (1/(y^2)-8/(y^4))(y+2y^3) ?

The derivative of (1/(y^2)-8/(y^4))(y+2y^3) is (15)/(y^2)+(24)/(y^4)+2

What is the first derivative of (1/(y^2)-8/(y^4))(y+2y^3) ?

The first derivative of (1/(y^2)-8/(y^4))(y+2y^3) is (15)/(y^2)+(24)/(y^4)+2

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derivative of

(\frac{1}{y ^{2}} - \frac{8}{y ^{4}}) (y + 2 y^{3})

\frac{15}{y ^{2}} + \frac{24}{y ^{4}} + 2

Solution steps

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{8}{y ^{4}}) (y + 2 y^{3}))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = \frac{1}{y ^{2}} - \frac{8}{y ^{4}}, g = y + 2 y^{3}

= \frac{d}{d y} (\frac{1}{y ^{2}} - \frac{8}{y ^{4}}) (y + 2 y^{3}) + \frac{d}{d y} (y + 2 y^{3}) (\frac{1}{y ^{2}} - \frac{8}{y ^{4}})

\frac{d}{d y} (\frac{1}{y ^{2}} - \frac{8}{y ^{4}}) = - \frac{2}{y ^{3}} + \frac{32}{y ^{5}}

\frac{d}{d y} (y + 2 y^{3}) = 1 + 6 y^{2}

= (- \frac{2}{y ^{3}} + \frac{32}{y ^{5}}) (y + 2 y^{3}) + (1 + 6 y^{2}) (\frac{1}{y ^{2}} - \frac{8}{y ^{4}})

Simplify

(- \frac{2}{y ^{3}} + \frac{32}{y ^{5}}) (y + 2 y^{3}) + (1 + 6 y^{2}) (\frac{1}{y ^{2}} - \frac{8}{y ^{4}}) : \frac{15}{y ^{2}} + \frac{24}{y ^{4}} + 2

= \frac{15}{y ^{2}} + \frac{24}{y ^{4}} + 2

\frac{5}{s ^{2} + 1}

\frac{d}{d x} ((x + \frac{1}{x ^{3}})^{2})

\int (\frac{x ^{2} + 1}{x ^{3} + 3 x}) d x

y = \frac{4 x ^{2} + 8 x + 4}{x}

x \to 5 lim (5 - x)

What is the derivative of (1/(y^2)-8/(y^4))(y+2y^3) ?
The derivative of (1/(y^2)-8/(y^4))(y+2y^3) is (15)/(y^2)+(24)/(y^4)+2
What is the first derivative of (1/(y^2)-8/(y^4))(y+2y^3) ?
The first derivative of (1/(y^2)-8/(y^4))(y+2y^3) is (15)/(y^2)+(24)/(y^4)+2