Can anyone solve this Calculus problem: Find the first derivative for f(x)=cos^2(tan x)?

Answer 1

Notice that the above answers are giving you two sets of answers, both of which are correct.

f'(x) = -2*sin(tanx)*cos(tanx) * sec^2 (x)
which can also be written as
-sin(2tanx) * sec^2 (x)

Answer 2

tan x = T

f(x) = cos^2 T

df/dt = 2 cos T DT(cost ) = - 2 cos T sin T

dt/dx = sec^ 2 x

df/dx = DT/dt*dt/dx = -2 cos t sin t sec^2 x
= - 2 cos (tan x) sin (tan x) sec^2 x

Answer 3

a. first, remedy for y: y=(one hundred-x^2)^(a million/2) then exchange something for the one hundred-x^2=U: y=U^(a million/2) then derivatize that: .5du*U^(-a million/2) and the one hundred-x^2: -2x=dU then placed U and dU interior the equation: dy/dx=.5(-2x)(one hundred-x^2)^(-a million/2) put in 6 for x and the slope is -3/4 b. even greater straightforward, derivatise the equation: dy/dx=2x put in -a million for x to get the slope: -2 and make a line with a slope of -2 that is going with the aid of -a million,a million the equation for the line is: y=-2x-a million

Answer 4

let tan x = t

f(x) = cos^2 t

df/dt = 2 cos t dt(cost ) = - 2 cos t sin t

dt/dx = sec^ 2 x

df/dx = dt/dt*dt/dx = -2 cos t sin t sec^2 x
= - 2 cos (tan x) sin (tan x) sec^2 x

Answer 5

y = cos² (tan x)
Let u = tan x
du/dx = sec² x
y = (cos u)²
dy/du = - 2 sin u.cos u
f `(x) = dy/dx = - 2 sin (tan x).cos (tan x).sec² x

Answer 6

y=f(x)
y=cos^2(tan x)
u=tan x
y=cos^2 u
dy/du = -2cos u sin u
= -sin(2u)
= -sin(2tan x)

du/dx = sec^2 (x)
dy/dx = dy/du * du/dx = -sec^2(x).sin(2tanx)

f'(x) = -sec^2(x).sin(2tanx)

Answer 7

Mental substitution approach:
f'(x) = 2cos(tanx)(-sin(tanx))(sec^2x) = -sin(2tanx)sec^2x