find derivative?

Question

(e^-5x)(cos^-1(2x))

Answer 1

Hi,
(e^-5x)(cos^-1(2x))
Use the product rule together with arccos (2x) which is:-2/(√(1-4x²).
e^(-5x) (-2/(√(1-4x²))+ cos^(-1) (2x) (-5e^(-5x))
e^(-5x)[ (-2/(√(1-4x²)) -5 cos^(-1) (2x)]
You may need to do some simplifying to get the answer in the form you want, but that’s the basic answer.

FE

Answer 2

To make things simpler, observe that cos^-1(2x) = sec(2x). So,

f(x) = e^(-5x) sec(2x). Using the product rule and the chain rule, we get

f'(x) = -5 e^(-5x) sec(2x) + e^(-5x) sec(2x) tan(2x) (2) =
f'(x) = -5 e^(-5x) sec(2x) + 2 e^(-5x) sec(2x) tan(2x)
f'(x) = e^(-5x) sec(2x) [2 tan(2x) - 5]

EDIT: I take it that by cos^(-1) you mean 1/cos and not arccos! It seems the persons who answered before me took it as arccos. If so, thei answers are OK

Answer 3

Use the product rule and chain rule.

y' = (-5)(e^-5x)(cos^-1 (2x) + (e^-5x)[-1 / sqrt(1 - (2x)^2)](2)

Answer 4

e^(-5x)[cos^-1(2x)]

let u = e^(-5x)

du = e^(-5x)(-5) = -5 e^(-5x)

v = cos^-1(2x)

dv = [-1/sqrt(1- 4x^2)](2) = -2/sqrt(1-4x^2)

d(uv) = udv + v du

= e^(-5x)[-2/sqrt(1-4x^2) + cos^-1(2x)[-5 e^(-5x)]

- e^(-5x) [(2/sqrt(1-4x^2)) + 5 cos^-1(2x)]