What is the derivative of y=((2x+1)(x^3+3))/x ?

The derivative of y=((2x+1)(x^3+3))/x is (6x^4+2x^3-3)/(x^2)

What is the first derivative of y=((2x+1)(x^3+3))/x ?

The first derivative of y=((2x+1)(x^3+3))/x is (6x^4+2x^3-3)/(x^2)

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derivative of

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

\frac{6 x ^{4} + 2 x ^{3} - 3}{x ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} ((2 x + 1) (x^{3} + 3)) = 8 x^{3} + 3 x^{2} + 6

\frac{d x}{d x} = 1

= \frac{( 8 x ^{3} + 3 x ^{2} + 6 ) x - 1 \cdot ( 2 x + 1 ) ( x ^{3} + 3 )}{x ^{2}}

Simplify

\frac{( 8 x ^{3} + 3 x ^{2} + 6 ) x - 1 \cdot ( 2 x + 1 ) ( x ^{3} + 3 )}{x ^{2}} : \frac{6 x ^{4} + 2 x ^{3} - 3}{x ^{2}}

= \frac{6 x ^{4} + 2 x ^{3} - 3}{x ^{2}}

4 x^{2} - 2 x + 3

\int_{0}^{π} sin (x) + cos (2 x) d x

y^{^{'}} + \frac{4}{5} y = 1 - \frac{1}{4} t, y (0) = 0

\int \frac{e ^{2 x}}{1 - e ^{x}} d x

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

What is the derivative of y=((2x+1)(x^3+3))/x ?
The derivative of y=((2x+1)(x^3+3))/x is (6x^4+2x^3-3)/(x^2)
What is the first derivative of y=((2x+1)(x^3+3))/x ?
The first derivative of y=((2x+1)(x^3+3))/x is (6x^4+2x^3-3)/(x^2)