derivative of (((-2x-1)/(-4x+2))^5)

⚠ 2x is read as the product 2*x: for a power, write 2^x

⚠ 4x is read as the product 4*x: for a power, write 4^x

1) Let's differentiate:

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

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

Apply the differentiation rule:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{(n - 1)}

with f =

- \frac{2 x + 1}{- 4 x + 2}

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

For

u1

2 x + 1

and

v2

- 4 x + 2

, use the product/quotient rule:

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

2) Let's perform the differentiation

u1

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

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

2

3) Let's perform the differentiation

v2

\frac{d}{d x} (- 4 x + 2)

\frac{d}{d x} (- 4 x) + 0

-4

4) You can evaluate the resulted derivative (

\frac{u1}{v2}

)':

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

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

ℹ

2 \cdot (- 4 x + 2)

- 8 x + 4

‖

(2 x + 1) \cdot 4

8 x + 4

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

ℹ

4 + 4

8

‖

- 8 x + 8 x

0

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

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

5) Then you can continue the derivation:

5 \cdot (- \frac{8}{{(- 4 x + 2)}^{2}}) \cdot {(\frac{2 x + 1}{- 4 x + 2})}^{4}

- \frac{40 \cdot {(\frac{2 x + 1}{- 4 x + 2})}^{4}}{{(- 4 x + 2)}^{2}}

- \frac{40 \cdot {(\frac{2 x + 1}{2 - 4 x})}^{4}}{{(2 - 4 x)}^{2}}

- \frac{40 \cdot \frac{{(2 x + 1)}^{4}}{{(2 - 4 x)}^{4}}}{{(2 - 4 x)}^{2}}

- \frac{\frac{40 \cdot {(2 x + 1)}^{4}}{{(2 - 4 x)}^{4}}}{{(2 - 4 x)}^{2}}

- \frac{40 \cdot {(2 x + 1)}^{4}}{{(2 - 4 x)}^{4}} \cdot \frac{1}{{(2 - 4 x)}^{2}}

▶Derivative:

- \frac{40 \cdot {(2 x + 1)}^{4}}{{(2 - 4 x)}^{6}}