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# are the differentials of sec^2x and tan^2x the same? watch

1. I worked out both as dy/dx = 2tanxsec^2x
2. (Original post by Abi1997)
I worked out both as dy/dx = 2tanxsec^2x
That is the secant
3. (Original post by Abi1997)
I worked out both as dy/dx = 2tanxsec^2x
You use the chain rule
Let y =(tanx)^2
Dy = 2tanxsec^2x
Y=sec^2x=tan^2x-1
Oh yeah your right
4. (Original post by Abi1997)
I worked out both as dy/dx = 2tanxsec^2x
Yeah that's right.
5. thanks for the speedy reply, much appreciated.
6. (Original post by Abi1997)
I worked out both as dy/dx = 2tanxsec^2x
dy/dx is not an operator. dy/dx is the derivative of y with respect to x which is d/dx(y). You can say dy/dx of tanx is (secx)^2. It doesn't make sense.
But yes the derivatives are the same.
7. (Original post by Abi1997)
thanks for the speedy reply, much appreciated.
sec squared and tan squared are identical up to a constant. So the derivatives are going to be equal.
8. Similar with sin squared and cos squared.

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