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

The derivative of (1/(y^2)-2/(y^4))(y+8y^3) is (15)/(y^2)+6/(y^4)+8

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

The first derivative of (1/(y^2)-2/(y^4))(y+8y^3) is (15)/(y^2)+6/(y^4)+8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

\frac{15}{y ^{2}} + \frac{6}{y ^{4}} + 8

Solution steps

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

Apply the Product Rule:

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

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

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

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

\frac{d}{d y} (y + 8 y^{3}) = 1 + 24 y^{2}

= (- \frac{2}{y ^{3}} + \frac{8}{y ^{5}}) (y + 8 y^{3}) + (1 + 24 y^{2}) (\frac{1}{y ^{2}} - \frac{2}{y ^{4}})

Simplify

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

= \frac{15}{y ^{2}} + \frac{6}{y ^{4}} + 8

\int_{- 2}^{3} \frac{35}{x ^{4}} d x

\frac{d y}{d x} = e^{2 x} + 7 y

\frac{d}{d x} (\frac{y}{x - y})

y = 2 x^{3} - x^{2} + 9, (2, 21)

\int \frac{1}{x - 4} d x

What is the derivative of (1/(y^2)-2/(y^4))(y+8y^3) ?
The derivative of (1/(y^2)-2/(y^4))(y+8y^3) is (15)/(y^2)+6/(y^4)+8
What is the first derivative of (1/(y^2)-2/(y^4))(y+8y^3) ?
The first derivative of (1/(y^2)-2/(y^4))(y+8y^3) is (15)/(y^2)+6/(y^4)+8