Can you solve this calculus problem?

In this video we solve another tricky derivative problem!

[jan-willemreens]

... Y = COS(X) ... [ COS(X) = SQRT(1 - SIN^2(X) ] ... Y = [ 1 - (SIN(X))^2 ]^(1/2) ... dY/d(SIN(X)) = (1/2) * [ 1 - (SIN(X))^2 ]^(- 1/2) * ( (- 2) * SIN(X) * d(SIN(X)/d(SIN(X) ) = - SIN(X) * 1/( [ COS^2(X) ]^(1/2) ) = - SIN(X)/COS(X) = - TAN(X) ... in case people get confused (lol) ... good to see you back Alex on YouTube with your instructive presentations " Alex Li " ... best regards, Jan-W

[richardbloemenkamp]

Actually if you do substitution you need the derivative of arcsin(u). If you forgot it a little then the method in the video is a check. u=sin(x) and then dcos(x)/dsin(x) = cos(arcsin(u))'= -u/sqrt(1-u^2) = -tan(x) works without issue.

[randum]

If this is Oxford
Then I should be in MIT

[stephenlesliebrown]

Before watching I thought, "try letting u = sin x and see what happens." Well, it ended up a moderately large expression involving sin and sin squared of x which does reduce to -tan x, but the way to do the problem is definitely by the chain rule!!
Best wishes to all 🙂

[matthewfeig]

Start from (sin x)^2+(cos x)^2=1 and take the derivative with respect to sin x. 2s +2c*dc/ds=0.
Solve for the derivative to get dc/ds = -s/c = -tan x.

[renesperb]

I would do it like this : set t= sinx . Then √(1-t^2) ' = -t/√(1-t^2) = -sin x/cos x = -tan x .

[cheesebusiness]

What is “differentiate A in respect to B” at the first place?

[ganesh]

When you started by conveying the Chain rule, I thought you would use the Chain rule, and do (d(cox) /dx) (dx / d(sinx))