What is the derivative of (3y^2)/(4-15xy) ?

The derivative of (3y^2)/(4-15xy) is (3(8y-15xy^2))/((4-15xy)^2)

What is the first derivative of (3y^2)/(4-15xy) ?

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

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derivative of

\frac{3 y ^{2}}{4 - 15 x y}

\frac{3 ( 8 y - 15 x y ^{2} )}{( 4 - 15 x y ) ^{2}}

Solution steps

\frac{d}{d y} (\frac{3 y ^{2}}{4 - 15 x y})

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d y} (\frac{y ^{2}}{4 - 15 x y})

Apply the Quotient Rule:

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

= 3 \cdot \frac{\frac{d}{d y} ( y ^{2} ) ( 4 - 15 x y ) - \frac{d}{d y} ( 4 - 15 x y ) y ^{2}}{( 4 - 15 x y ) ^{2}}

\frac{d}{d y} (y^{2}) = 2 y

\frac{d}{d y} (4 - 15 x y) = - 15 x

= 3 \cdot \frac{2 y ( 4 - 15 x y ) - ( - 15 x ) y ^{2}}{( 4 - 15 x y ) ^{2}}

Simplify

3 \cdot \frac{2 y ( 4 - 15 x y ) - ( - 15 x ) y ^{2}}{( 4 - 15 x y ) ^{2}} : \frac{3 ( 8 y - 15 x y ^{2} )}{( 4 - 15 x y ) ^{2}}

= \frac{3 ( 8 y - 15 x y ^{2} )}{( 4 - 15 x y ) ^{2}}

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

n = 0 \sum \infty \frac{( 2 ^{n + 1} )}{5 ^{n}}

\int \frac{x}{5 x + 3} d x

\int \frac{1}{100} x + 10 d x

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