What is the (arctan(4x))^' ?

The (arctan(4x))^' is 4/(16x^2+1)

What is the first (arctan(4x))^' ?

The first (arctan(4x))^' is 4/(16x^2+1)

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(arctan (4 x))^{^{'}}

\frac{4}{16 x ^{2} + 1}

Solution steps

(arctan (4 x))^{'}

Apply the chain rule:

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

= \frac{1}{( 4 x ) ^{2} + 1} (4 x)^{'}

(4 x)^{'} = 4

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

Simplify

\frac{1}{( 4 x ) ^{2} + 1} \cdot 4 : \frac{4}{16 x ^{2} + 1}

= \frac{4}{16 x ^{2} + 1}

\int \frac{5 x ^{2}}{4 x - x ^{2}} d x

\int \frac{4}{u} d u

n = 2 \sum \infty e^{5 - 2 n}

\int - 16 x d x

\int (x + 1)^{10} (x + 2) d x