What is the derivative of (5x^2+4/(x+3)) ?

The derivative of (5x^2+4/(x+3)) is (5x^2+30x-4)/((x+3)^2)

What is the first derivative of (5x^2+4/(x+3)) ?

The first derivative of (5x^2+4/(x+3)) is (5x^2+30x-4)/((x+3)^2)

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\frac{d}{d x} (\frac{5 x ^{2} + 4}{x + 3})

\frac{5 x ^{2} + 30 x - 4}{( x + 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{5 x ^{2} + 4}{x + 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 5 x ^{2} + 4 ) ( x + 3 ) - \frac{d}{d x} ( x + 3 ) ( 5 x ^{2} + 4 )}{( x + 3 ) ^{2}}

\frac{d}{d x} (5 x^{2} + 4) = 10 x

\frac{d}{d x} (x + 3) = 1

= \frac{10 x ( x + 3 ) - 1 \cdot ( 5 x ^{2} + 4 )}{( x + 3 ) ^{2}}

Simplify

\frac{10 x ( x + 3 ) - 1 \cdot ( 5 x ^{2} + 4 )}{( x + 3 ) ^{2}} : \frac{5 x ^{2} + 30 x - 4}{( x + 3 ) ^{2}}

= \frac{5 x ^{2} + 30 x - 4}{( x + 3 ) ^{2}}

x \to - 1 lim (\frac{( 1 - x )}{x + 1})

y = 3 6 x^{2} + 1

- 3^{- 2} (tan (7 t))^{5}

x \to 3 lim (\frac{\frac{x}{3} + 2}{x + 1})

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

What is the derivative of (5x^2+4/(x+3)) ?
The derivative of (5x^2+4/(x+3)) is (5x^2+30x-4)/((x+3)^2)
What is the first derivative of (5x^2+4/(x+3)) ?
The first derivative of (5x^2+4/(x+3)) is (5x^2+30x-4)/((x+3)^2)