Calculus, differentiate?

Question

Differentiate the function y = cos(ln(x+1))

Answer 1

y = cos(ln(x+1))

let t = ln (x+1)

y = cos t
dy/dx = dy/dt. dt/dx
= - sin t . 1/(x+1) using chain rule

= - (sin (ln(x+1)))(x+1)

Answer 2

Let g(x) = ln(x+1)
Therefore:
y = cos(g(x))
y' = -g'(x)*sin(g(x))

Finding g'(x):
g(x) = ln(x+1)
g'(x) = 1 / (x+1)

Therefore:
y' = -cos(ln(x+1)) / (x + 1)

Answer 3

chain rule:
-sin(ln(x+1))/(x+1);
derivative of ln = 1/(x+1)
derivative of cos(ln(x+1)) = -sin(ln(x+1)) x derivative of ln

Answer 4

y = cos(ln(x+1))
Let u = ln(x+1)
Then dy/dx = -sinu du/dx
= -sin(ln(x+1))(1/(x+1))

Answer 5

use the chain rule. take the spinoff of the exterior functionality at the same time as leaving the interior on my own and then take the spinoff of the interior functionality. spinoff of ln(x) = a million/x, so spinoff of ln(3x-4) is a million/(3x-4) * 3 = 3 / (3x-4)

Answer 6

did you even try. What was your guess?