What is the d/(dy)(1/(y^3+4y^2)) ?

The d/(dy)(1/(y^3+4y^2)) is -(3y+8)/(y^3(y+4)^2)

What is the first d/(dy)(1/(y^3+4y^2)) ?

The first d/(dy)(1/(y^3+4y^2)) is -(3y+8)/(y^3(y+4)^2)

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

What is the d/(dy)(1/(y^3+4y^2)) ?
The d/(dy)(1/(y^3+4y^2)) is -(3y+8)/(y^3(y+4)^2)
What is the first d/(dy)(1/(y^3+4y^2)) ?
The first d/(dy)(1/(y^3+4y^2)) is -(3y+8)/(y^3(y+4)^2)