What is the d/(dt)(arctan(4t)) ?

The d/(dt)(arctan(4t)) is 4/(16t^2+1)

What is the first d/(dt)(arctan(4t)) ?

The first d/(dt)(arctan(4t)) is 4/(16t^2+1)

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\frac{d}{d t} (arctan (4 t))

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

Solution steps

\frac{d}{d t} (arctan (4 t))

Apply the chain rule:

\frac{1}{( 4 t ) ^{2} + 1} \frac{d}{d t} (4 t)

= \frac{1}{( 4 t ) ^{2} + 1} \frac{d}{d t} (4 t)

\frac{d}{d t} (4 t) = 4

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

Simplify

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

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

\int cos (t) - 5 sin (t) d t

x \to 0 + lim (x ln (tan (5 x)))

4 s

f (x) = \frac{2}{x}, at x = - 9

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

What is the d/(dt)(arctan(4t)) ?
The d/(dt)(arctan(4t)) is 4/(16t^2+1)
What is the first d/(dt)(arctan(4t)) ?
The first d/(dt)(arctan(4t)) is 4/(16t^2+1)