- #1
andrey21
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Find the derivative of the following:
y = 1/ COS(t^2)
Here is my attempt
Using (Cos(t^2))^-1
SOlve using the chain rule:
u = t^2
du = 2t
dy/dt = -1.(-sin(u))^(-2) .(2t)
= -2t(-sin(u))^(-2)
= -2t/(Sin(t^2))^2
Is this correct or have I totally used the wrong techiniques?
y = 1/ COS(t^2)
Homework Equations
Here is my attempt
Using (Cos(t^2))^-1
SOlve using the chain rule:
u = t^2
du = 2t
dy/dt = -1.(-sin(u))^(-2) .(2t)
= -2t(-sin(u))^(-2)
= -2t/(Sin(t^2))^2
Is this correct or have I totally used the wrong techiniques?