answersLogoWhite

0


Best Answer

First find the derivative of each term. The derivative of any constant is zero, so d(1)/dx=0. To find the derivative of cos2x, use the chain rule.

d(cos2x)/dx=-sin(2x)(2)=-2sin(2x)

So the answer is 0-2sinx, or simply -2sinx

User Avatar

Wiki User

โˆ™ 2010-03-04 21:45:28
This answer is:
User Avatar
Study guides

Algebra

20 cards

A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

โžก๏ธ
See all cards
3.74
โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…โ˜†โ˜…
818 Reviews

Add your answer:

Earn +20 pts
Q: Derivative of 1 plus cos2x
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

What is the Derivative of -x plus tanx?

(-x+tanx)'=-1+(1/cos2x)


Derivative of tanx?

It is sec2x, this is the same as 1/cos2x.


How do you prove one - tan square x divided by one plus tan square xequal to cos two x?

(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)


How do you verify cos2 theta divided by csc2 theta plus cos4 theta equals cos2 theta?

Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x


How does sin2x divided by 1-cosx equal 1 plus cosx?

sin2x / (1-cos x) = (1-cos2x) / (1-cos x) = (1-cos x)(1+cos x) / (1-cos x) = (1+cos x) sin2x=1-cos2x as sin2x+cos2x=1 1-cos2x = (1-cos x)(1+cos x) as a2-b2=(a-b)(a+b)


How do you factor sinx-cos2x-1?

[sinx - cos2x - 1] is already factored the most it can be


What are the identities in trigonometry?

Sin2x + Cos2x=1, so Cos2x=1-Sin2x and Sin2x=1-Cos2x. Also Sin/Cos = Tan. Sec2x=1+Tan2x. Cot2x+1=Csc2x.


Integral of sin squared x?

First we look at the double-angle identity of cos2x. We know that: cos2x = cos^2x - sin^2x cos2x = [1-sin^2x] - sin^2x.............. (From sin^2x + cos^2x = 1, cos^2x = 1 - sin^2x) Therefore: cos2x = 1 - 2sin^2x 2sin^2x = 1 - cos2x sin^2x = 1/2(1-cos2x) sin^2x = 1/2 - cos2x/2 And intergrating, we get: x/2 - sin2x/4 + c...................(Integral of cos2x = 1/2sin2x; and c is a constant)


How do you solve 2 cos squared x - sinx - 1?

1


How does 1 plus cot squared x equals csc squared x?

The proof of this trig identity relies on the pythagorean trig identity, the most famous trig identity of all time: sin2x + cos2x = 1, or 1 - cos2x = sin2x. 1 + cot2x = csc2x 1 = csc2x - cot2x 1 = 1/sin2x - cos2x/sin2x 1 = (1 - cos2x)/sin2x ...using the pythagorean trig identity... 1 = sin2x/sin2x 1 = 1 So this is less of a proof and more of a verification.


What is the derivative of ln 1 plus x?

d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)


Integral of cos2x log cosx-sinx coax plus since?

it is not possible to get the Integral of cos2x log cosx-sinx coax plus since there are no symbols given in the equation.

People also asked