The Student Room Group
you look sec^2 up in the formula book (it should be there) then you need to do the product rule. and that should give you an answer -i hope-
or possibly the chain rule thing I think it is
Reply 3
No sec2sec^2 isn't in the formula book, only the integral of it - tanx
:s-smilie:
Reply 4
If you have any function, then adding a constant within that function (if you know what I mean) doesn't make any difference. For example: (c is a constant)
ddxsin(x+c)=cos(x+c)\frac{d}{dx} \sin (x + c) = \cos (x + c)
ddx(x+c)3=3(x+c)2\frac{d}{dx} (x + c)^3 = 3(x + c)^2
In general, if ddxf(x)=f(x)\frac{d}{dx} f(x) = f'(x), then ddxf(x+c)=f(x+c)\frac{d}{dx} f(x + c) = f'(x + c). You can apply this logic to the differentiation you're trying to do.
Reply 5
Yes i know, i can't integrate
sec2(x)sec^2(x) either :frown:
Reply 6
If you're stuck on differentiating sec^2 x, then use chain rule, the derivative of sec x is in the formula book.
EDIT: above post nulls suggestion :smile:
eipi
Yes i know, i can't integrate
sec2(x)sec^2(x) either :frown:


Have you managed to differentiate?

Unparseable latex formula:

\sec^2{x} = (\sec{x}})^2



The derivative of secx is secx tanx. Use the chain rule.

To integrate, sec2x \sec^2{x} is the derivative of tanx... (it's in the formula book)
yup remember sec^2 = (cos x)^-2

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