whats the derivative using the chain rule for cos(x+a)?

Question

how about t^2(csc(t)?

Answer 1

Let u = x + a
du/dx = 1
y = cos u
dy/du = - sin u = - sin (x + a)
dy / dx = (dy / du) . (du / dx)
dy / dx = - sin ( x + a )

Answer 2

Like cos(x), but with x+a, because x and x+a share the derivative 1.


d/dx cos(x+a) = -sin(x + a)

Answer 3

set u= sin^3x and v= tan4x then calculate the by-product of each so u' = 3sin^2xcosx and v'= 4sec^2(4x) chain rule provides uv'+vu' so that you get (sin^3x)(4sec^2(4x))+ (tan(4x))(3sin^2xcosx) which will properly be simplified to sin^2(x) (3 cos(x) tan(4 x)+4 sin(x) sec^2(4 x))

Answer 4

If a is simply a number and not a variable, then the derivitive is:
-sin(x+a)

let u=x+a
du=dx

so,
d/dx(cos(u))=-sin(u)
=-sin(x+a)

Answer 5

cos(x+a) *assuming a is a constant*

derivative of f(g(x)) = f'(g(x))*g'(x)

-sin(x+a)*(1)

-sin(x+a)

Answer 6

let y = x + a
we get
d/dx(cos y) = -sin y dy/dx = - sin y . 1 or - sin y = - sin(x+a)

Answer 7

differentiate cos(x+a) with respect to (x+a) and differentiate (x+a) with respect to x. thats it
the answer is -sin(x+a) assuming "a" is a constant.

Answer 8

o= cosx i= x+a
o'= -sinx i'= 1

-sin(x+a) * 1

Answer 9

f(x)=cos(x) ===> f'(x)= -sin(x)
g(x)=x+a ===> g'(x) = 1

your function h(x)=cos(x+a)=cos(g(x))=f(g(x))

h'(x)=g'(x)f'(g(x)) = (1) [-sin(x+a)] = -sin(x+a)