What is the derivative of ((4-80y)/y)(y-48) ?

The derivative of ((4-80y)/y)(y-48) is (-80y^2+192)/(y^2)

What is the first derivative of ((4-80y)/y)(y-48) ?

The first derivative of ((4-80y)/y)(y-48) is (-80y^2+192)/(y^2)

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derivative of

(\frac{4 - 80 y}{y}) (y - 48)

\frac{- 80 y ^{2} + 192}{y ^{2}}

Solution steps

\frac{d}{d y} (\frac{4 - 80 y}{y} (y - 48))

Apply the Product Rule:

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

f = \frac{4 - 80 y}{y}, g = y - 48

= \frac{d}{d y} (\frac{4 - 80 y}{y}) (y - 48) + \frac{d}{d y} (y - 48) \frac{4 - 80 y}{y}

\frac{d}{d y} (\frac{4 - 80 y}{y}) = - \frac{4}{y ^{2}}

\frac{d}{d y} (y - 48) = 1

= (- \frac{4}{y ^{2}}) (y - 48) + 1 \cdot \frac{4 - 80 y}{y}

Simplify

(- \frac{4}{y ^{2}}) (y - 48) + 1 \cdot \frac{4 - 80 y}{y} : \frac{- 80 y ^{2} + 192}{y ^{2}}

= \frac{- 80 y ^{2} + 192}{y ^{2}}

\frac{\partial}{\partial x} (f (x))

\int \frac{1}{x ^{2} - 4 x + 5} d x

\int 8 x - x^{2} d x

\int x 12 - x^{2} d x

y = 1 - x^{2}, y = 4 - 4 x

What is the derivative of ((4-80y)/y)(y-48) ?
The derivative of ((4-80y)/y)(y-48) is (-80y^2+192)/(y^2)
What is the first derivative of ((4-80y)/y)(y-48) ?
The first derivative of ((4-80y)/y)(y-48) is (-80y^2+192)/(y^2)