derivative of arccot(sqrt(2x)) — step-by-step solution

⚠ arccot(sqrt(2x)) is read as (pi/2-atan(sqrt(2x))): for the real arccot branch with values in (0, π), use pi/2-atan(expression)‖2x is read as the product 2*x: for a power, write 2^x

Let's perform the differentiation:

\frac{d}{d x} (\frac{π}{2} - atan (\sqrt{2 x}))

\frac{d}{d x} (\frac{π}{2}) + \frac{d}{d x} (- atan (\sqrt{2 x}))

We differentiate using:

\frac{d}{d x} (\arctan (f)) = \frac{d}{d x} (f) \cdot \frac{1}{f^{2} + 1}

(where f =

\sqrt{2 x}

)

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

Apply the differentiation rule:

\frac{d}{d x} (\sqrt{f}) = \frac{d}{d x} (f) \cdot \frac{1}{2 \sqrt{f}}

with f =

2 x

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

ℹ

\frac{d}{d x} (2 x) \cdot \frac{1}{2 \cdot \sqrt{2 x}}

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

‖

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

2

‖

\frac{1}{2} \cdot 2

\frac{2}{2}

‖

\frac{2}{2}

1

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

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

- \frac{1}{\sqrt{2 x}} \cdot \frac{1}{2 x + 1}

▶Derivative:

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