What is (\partial)/(\partial y)((2x+3)/(4y+1)) ?

The answer to (\partial)/(\partial y)((2x+3)/(4y+1)) is -(4(2x+3))/((4y+1)^2)

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(\partial)/(\partial y)((2x+3)/(4y+1))

\frac{\partial}{\partial y} (\frac{2 x + 3}{4 y + 1})

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

Solution steps

\frac{\partial}{\partial y} (\frac{2 x + 3}{4 y + 1})

Treat

x

as a constant

Take the constant out:

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

= (2 x + 3) \frac{\partial}{\partial y} (\frac{1}{4 y + 1})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial y} (4 y + 1) = 4

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

Simplify

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

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

\int x e^{x} d x

y^{^{''}} - \frac{y ^{^{'}}}{x} = 3 x

\frac{x ^{5}}{5}

y^{^{'}} + 2 y = x e^{- 2 x}

(4 x - 2)^{3} (x^{2} + 1)^{2}

What is (\partial)/(\partial y)((2x+3)/(4y+1)) ?
The answer to (\partial)/(\partial y)((2x+3)/(4y+1)) is -(4(2x+3))/((4y+1)^2)