What is the (arctan(1/x))^' ?

The (arctan(1/x))^' is -1/(1+x^2)

What is the first (arctan(1/x))^' ?

The first (arctan(1/x))^' is -1/(1+x^2)

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(arctan (\frac{1}{x}))^{^{'}}

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

Solution steps

(arctan (\frac{1}{x}))^{'}

Apply the chain rule:

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

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

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

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

Simplify

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

= - \frac{1}{1 + x ^{2}}

\frac{\partial}{\partial x} (x^{4 c o s (x)})

\int \frac{4 + x ^{2}}{x ^{3}} d x

\frac{d Q}{d t} = 3 - \frac{3 Q}{1000}

f (x) = 4^{s i n^{2} (x) + c o s (3 x)}

x \to - 1 lim (∣ x ∣ - x)