1/sec(x)= cos(x). What is the second derivative of cos(x)?

Either write 2/cos(x)= 2sec(x) or "remember" the derivative of cos(x) or use the quotient rule: [tex]\frac{0(cos(x))- 2(-sin(x))}{cos^2(x)}= 2\frac{sin(x)}{cos(x)}\frac{1}{cos(x)}= 2tan(x)sec(x)2)2/cosx

1/tan(x)= cot(x)= cos(x)/sin(x). Either "remember" the derivative of cot(x) or use the quotient rule:3) 1/tanx

sec(x)= 1/cos(x) so 1/sec(2x)= cos(2x). Its derivative, by the chain rule, is -2sin(2x)4)1/sec2x

my answer is :

1) 1/cosx

2)2/-sinx

3)1/sex2x

4)1/cos2x[/quote]

Every one of these is wrong. You need more practice! And remember that the derivative of f(x)/g(x) is NOT f'(x)/g'(x)! It is

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By "ud(2)" do you mean the derivative of u at x= 2, u'(2)= 3?B) suppose that u is a differntiable function of x and that u(2) = 2 ud(2) =3

the derivaives of (2 + u ) respect with x . at x = 2 is ......(4)........

You should have learned long ago that the derivative of f(x)+ g(x) is f'(x)+ g'(x) and that the derivative of a constant is 0: (2+ u)'= u' and since u'(2)= 3, the derivative of 2+ u at x= 2 is also 3.

[quote]c) suppose that u is a differntiable function of x and that u(2) = 2 ud(2) =3

the derivaives of (2 + u ) respect with x . at x = 3 is ......(5)........Once again, the derivative of 2+ u is just the derivative of u since the derivative of the constant, 2, is 0. Since you do not know the derivative of uat x= 3you cannot say anything about the derivative of 2+ u at x= 3. Are you sure you have copied these problems correctly?

I want your help .