Who knows derivatives of Trig Functions?

Question

I am going over a sample calc quiz for Thursday and one of the problems is
y= sin{cos^2(tanx)} i started the problem by using the chain rule right and i got
dy/dx= cos{cos^2(tanx)} * ?
i know your supposed to multiply iit by dy/dx of the inside of the parenthesis so my question is what is the derivative of cos^2(tanx)

Answer 1

y = cos^2(tan(x))
dy/dx = 2cos(tan(x)) (d/dx)(cos(tan(x))
= 2cos(tan(x)) (-sin(tan(x))) (d/dx)(tan(x))
= - 2 cos(tan(x)) sin(tan(x)) sec^2(x).

That is the answer to your question 'what is the derivative of cos^2(tanx)?'. Going back to the sample quiz question, of course, this answer has to be multiplied by cos{cos^2(tanx)} as you say.

Answer 2

you have to apply chain rule several times

Let u = tanx du/dx = sec^2 x

d(cos^2(u))/ du = -2cos(u) sin(u)

so dy/dx = cos{cos^2(tanx)} (-2(cos(tanx)sin(tanx)) sec^2 x

Answer 3

the derivative of cos^2(tanx) is 2cos(tanx)sin(tanx) by power and chain rules, times the derivative of tanx which is sec^2x, so lets put all that together, this derivative of cos^2(tanx) is 2cos(tanx)sin(tanx)sec^2x

now the derivative of the entire function is:

-2cos(cos^2(tanx))cos(tanx)sin(tanx)sec^2x

Answer 4

dy/dx
= cos{cos^2(tanx)}
[2cos(tanx)][-sin(tanx)]
sec^2(x)
= -cos{cos^2(tanx)}
[sin(2tanx)]sec^2(x)

For your question:
d/dx [cos^2(tanx)] = -[sin(2tanx)]sec^2(x)
--------
Ideas: Use chain rule. y = sinw, w = (cosu)^2, u = tanx
dy/dx = (dy/dw)(dw/du)(du/dx)

Answer 5

you will would desire to apply the chain rule. to illustrate in case you had to locate the by-fabricated from y = sin(x^2) it may be y' = cos(x^2)*2x = 2xcos(x^2). on your case i'm uncertain in case you propose pi/(3x) or (pi/3)x, it could make a difference interior the respond. relatively the chain rule says to locate the by-fabricated from the outdoors function evaluated on the interior function circumstances the by-fabricated from the interior function.

Answer 6

y = sin{cos²(tanx)}

u = cos²(tan(x))

y = sin(u)
dy/du = cos(u)
= cos(cos²(tan(x)))

u = cos²(tan(x))
Let v = tan(x)
u = cos²v
du/dv = -2sin(v)cos(v)
= -sin(2v)
= -sin(2tan(x))

v = tan(x)
dv/dx = sec²(x)

dy/dx = dy/du * du/dv * dv/dx
= cos(cos²(tan(x))) * (-sin(2tan(x))) * sec²(x)
= -sec²(x)cos(cos²(tan(x)))(sin(2tan(x)))