I need help with the chain rule.?

Question

Differentiate:

cos(sin(e^(x^2)))

Answer 1

Chain rule: dy/dx = dy/du * du/dx
-sin(sin(e^(x^2))) * cos(e^(x^2)) * e^(x^2) * 2x

Answer 2

Hint: Don't use the formula... My math teacher said that it makes it to hard. And I did the math and got the same answer (written differently but same number - for example x= 2@y instead of 2y times @ (@ = a randomly chosen variable)

f(x)= cos(sin(e^(x^2))) Let u = sin(e^(x^2)
f(x)= cos u
f'(x)= -sin u

Can't do anything until you break down u more


Let u = sin (e^(x^2)) Let v = (e^(x^2)) Let w = x^2
u= sin v v= e^w w= x^2

Start with w and take the derivative

dw/dx = 2x change of w over the change of x
dw = 2x(dx)

take the derivative of v
dv/dw = e^w
dv= (e^w)(dw)

Now plug in
dv = (e^(x^2))(2x(dx))

Finally, take the derivative of u and plug in
u = sin v
du/dv = cos v
du = dv cos v
du = (e^(x^2))(2x(dx)) cos (e^(x^2))


Now back to the original part of the problem

f'(x)= -sin u or dy/du = -sin u change of y over change of u
dy = -du sin u
dy = -(e^(x^2))(2x(dx)) cos (e^(x^2)) sin (sin(e^(x^2))
dy/dx = -(e^(x^2))(2x) cos (e^(x^2)) sin (sin(e^(x^2))

i divided both sides by dx (ie 2x times dx)...

If this seems longer than the formula, by all means use the formula, but this way you wont forget how to do... and if your teacher does not require simplification, then this would be an OK answer...

Answer 3

Start from the outside and work your way in:

-sin(sin(e^(x^2))) * d/dx[sin(e^(x^2))]
= -sin(sin(e^(x^2))) * cos(e^(x^2)) * d/dx[e^(x^2)]
= -sin(sin(e^(x^2))) * cos(e^(x^2)) * e^(x^2) * d/dx[x^2]
= -sin(sin(e^(x^2))) * cos(e^(x^2)) * e^(x^2) * (2x)

Answer 4

let y = cos(sin(e^(x²)))
= cosp where p = sinq where q = e^r where r = x²

(So p = sin(e^(x²)), q =e^(x²) and we already know r = x²)

Then dy/dx = dy/dp . dp/dq . dq/dr . dr/dx
= -sinp , cosq .e^r .2x
= -2x.e^(x²).cos( e^(x²)).sin (sin ( e^(x^2)))

(Writing it is more compkex that the process!!!)