what formula do you use 2 differentiate arctan(2x)?

Answer 1

You use the chain rule, because there is a function within a function. The chain rule is the derivative of the outside function, keeping the original inside function, multiplied by the derivative of the inside function. So,
d/dx arctan(x) = 1/ (1+x^2)
d/dx 2x = 2
d/dx arctan(2x) = 2 *[1/ (1+(2x)^2)]

Answer 2

This Site Might Help You.

RE:
what formula do you use 2 differentiate arctan(2x)?

Answer 3

Derivative Of Arctan 4x

Answer 4

d/dx [arctan u] = u'/(1+u^2)

where u' is the derivative of U

hence

d/dx arctan(2x) = 2/(1+4x^2)

I thin....it has been a long time. Check the rule out in any text or on the web, but I think I am right

Answer 5

Let y = arctan (2x) , then
2x = tan y

Differentiate both side,
2dx = sek^2 y dy
2dx = (1+tan^2 y) dy
dy/dx = 2 / (1+tan^2 y)
dy/dx = 2 / [1+(2x)^2 ]
dy/dx = 2 / (1+4x^2 )

Answer 6

arctan(2x) is an example of chain rule.

example f(g(x)) = g'(x)f'(g(x))

let u = g(x)

(d/du)(arctan(u)) = 1/(u^2 + 1)

and (d/dx)(2x) = 2

so we get 2/(4x^2 +1) for the answer

Answer 7

Let = 2x
dydx = 1/(1+u^2) du/dx
= 1/(1-4x^2) * 2
= 2/(1-4x^2).

In general d/dx(arctan u) = 1/(1+u^2) *du/dx, where u = f(x).

Answer 8

For the best answers, search on this site https://shorturl.im/hsXNN

y= x / (2 + 3ln(x)) Simply use the quotient rule. y' = [(2 + 3ln(x)) - x(3/x)] / (2 + 3ln(x))^2 y' = [2 + 3ln(x) - 3] / (2 + 3ln(x))^2 y' = [3ln(x) - 1] / (2 + 3ln(x))^2