What is the derivative of (5x^2/(x+47)) ?

The derivative of (5x^2/(x+47)) is (5(x^2+94x))/((x+47)^2)

What is the first derivative of (5x^2/(x+47)) ?

The first derivative of (5x^2/(x+47)) is (5(x^2+94x))/((x+47)^2)

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\frac{d}{d x} (\frac{5 x ^{2}}{x + 47})

\frac{5 ( x ^{2} + 94 x )}{( x + 47 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{5 x ^{2}}{x + 47})

Take the constant out:

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

= 5 \frac{d}{d x} (\frac{x ^{2}}{x + 47})

Apply the Quotient Rule:

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

= 5 \cdot \frac{\frac{d}{d x} ( x ^{2} ) ( x + 47 ) - \frac{d}{d x} ( x + 47 ) x ^{2}}{( x + 47 ) ^{2}}

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

\frac{d}{d x} (x + 47) = 1

= 5 \cdot \frac{2 x ( x + 47 ) - 1 \cdot x ^{2}}{( x + 47 ) ^{2}}

Simplify

5 \cdot \frac{2 x ( x + 47 ) - 1 \cdot x ^{2}}{( x + 47 ) ^{2}} : \frac{5 ( x ^{2} + 94 x )}{( x + 47 ) ^{2}}

= \frac{5 ( x ^{2} + 94 x )}{( x + 47 ) ^{2}}

y^{^{'}} = x^{2} - y - 2, y (0) = - 1

\frac{\partial}{\partial y} (2 x y + y)

e^{2}

f (x) = - x^{2} + 3 x

\frac{d}{d x} (c x^{- 1})

What is the derivative of (5x^2/(x+47)) ?
The derivative of (5x^2/(x+47)) is (5(x^2+94x))/((x+47)^2)
What is the first derivative of (5x^2/(x+47)) ?
The first derivative of (5x^2/(x+47)) is (5(x^2+94x))/((x+47)^2)