What is the derivative of ((10y^3-9y^2+6y-10))/((2y+3)) ?

The derivative of ((10y^3-9y^2+6y-10))/((2y+3)) is (40y^3+72y^2-54y+38)/((2y+3)^2)

What is the first derivative of ((10y^3-9y^2+6y-10))/((2y+3)) ?

The first derivative of ((10y^3-9y^2+6y-10))/((2y+3)) is (40y^3+72y^2-54y+38)/((2y+3)^2)

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derivative of

\frac{( 10 y ^{3} - 9 y ^{2} + 6 y - 10 )}{( 2 y + 3 )}

\frac{40 y ^{3} + 72 y ^{2} - 54 y + 38}{( 2 y + 3 ) ^{2}}

Solution steps

\frac{d}{d y} (\frac{10 y ^{3} - 9 y ^{2} + 6 y - 10}{2 y + 3})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d y} ( 10 y ^{3} - 9 y ^{2} + 6 y - 10 ) ( 2 y + 3 ) - \frac{d}{d y} ( 2 y + 3 ) ( 10 y ^{3} - 9 y ^{2} + 6 y - 10 )}{( 2 y + 3 ) ^{2}}

\frac{d}{d y} (10 y^{3} - 9 y^{2} + 6 y - 10) = 30 y^{2} - 18 y + 6

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

= \frac{( 30 y ^{2} - 18 y + 6 ) ( 2 y + 3 ) - 2 ( 10 y ^{3} - 9 y ^{2} + 6 y - 10 )}{( 2 y + 3 ) ^{2}}

Expand

(30 y^{2} - 18 y + 6) (2 y + 3) - 2 (10 y^{3} - 9 y^{2} + 6 y - 10) : 40 y^{3} + 72 y^{2} - 54 y + 38

= \frac{40 y ^{3} + 72 y ^{2} - 54 y + 38}{( 2 y + 3 ) ^{2}}

y^{^{'}} = y (1 - 0.0008 y)

5 x^{^{''}} + 20 x^{^{'}} + 20 x = 28

\int sin^{3} (x) \cdot cos^{2} (x) d x

\frac{\partial}{\partial x} (5 x^{4} y^{- 4})

\int a^{2 x} d x

What is the derivative of ((10y^3-9y^2+6y-10))/((2y+3)) ?
The derivative of ((10y^3-9y^2+6y-10))/((2y+3)) is (40y^3+72y^2-54y+38)/((2y+3)^2)
What is the first derivative of ((10y^3-9y^2+6y-10))/((2y+3)) ?
The first derivative of ((10y^3-9y^2+6y-10))/((2y+3)) is (40y^3+72y^2-54y+38)/((2y+3)^2)