derivative of f(x)=(2x^2-x-1)/(2x+2) — step-by-step solution | Explain MAths

derivative of f(x)=(2x^2-x-1)/(2x+2)

We save the definition:

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

Let's differentiate:

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

We differentiate

u1

=

2 x^{2} - x - 1

and

v2

=

2 x + 2

using:

\frac{d}{d x} (\frac{u1}{v2}) = \frac{\frac{d}{d x} (u1) \cdot v2 - u1 \cdot \frac{d}{d x} (v2)}{{v2}^{2}}

Let's perform the differentiation

u1

:

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

\frac{d}{d x} (2 x^{2}) + \frac{d}{d x} (- x) + 0

ℹ

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

=

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

‖

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

=

2 x

‖

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

=

-1

2 \cdot 2 x - 1

4 x - 1

Let's perform the differentiation

v2

:

\frac{d}{d x} (2 x + 2)

\frac{d}{d x} (2 x) + 0

2

You can evaluate the resulted derivative (

\frac{u1}{v2}

)':

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

ℹ

(4 x - 1) \cdot (2 x + 2)

=

4 \cdot x \cdot 2 \cdot x

+ 4 x \cdot 2

- 1 \cdot 2 x

- 1 \cdot 2

‖

- (2 x^{2} - x - 1) \cdot 2

=

- (4 x^{2} - x \cdot 2 - 2)

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

ℹ

4 \cdot 2

=

8

‖

x \cdot x

=

x^{2}

‖

-1 \cdot 2 x

=

-2 x

‖

-1 \cdot 2

=

-2

‖

- (4 x^{2} - x \cdot 2 - 2)

=

- 4 x^{2} + x \cdot 2 + 2

\frac{8 x^{2} + 8 x - 2 x - 2 - 4 x^{2} + x \cdot 2 + 2}{{(2 x + 2)}^{2}}

ℹ

- 2 + 2

=

0

‖

- 2 x

+ 0

- 4 x^{2}

+ x \cdot 2

=

0 - 4 x^{2}

‖

8 x^{2} - 4 x^{2}

=

4 x^{2}

\frac{4 x^{2} + 8 x + 0}{{(2 x + 2)}^{2}}

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

▶Derivative:

\frac{x \cdot (x + 2)}{{(x + 1)}^{2}}