differentiation of 2cos(3x-4): which way is correct?

[Gokuson]

is the differentiation of 2cos(3x-4)=

1.6cos(3x-4)

2.-6sin(3x-4)


please help me and try explain it with step thanks

 

[HallsofIvy]

1) This is a problem in "Calculus", not "differential equations".
(That's for your information, not a criticism. If you have just started studying Calculus, we can't expect you to know what "differential equations" is!)

2) The derivative of cos(x), with respect to x, is - sin(x).

3) Using the "chain rule" to find the derivative of cos(3x- 1), you will have to multiply -sin(3x- 4) times the derivative of 3x- 4 which is 3.

 

[stapel]

Topic moved from "Differential Equations" to "Calculus".

 

[srmichael]

1) This is a problem in "Calculus", not "differential equations".
(That's for your information, not a criticism. If you have just started studying Calculus, we can't expect you to know what "differential equations" is!)

2) The derivative of cos(x), with respect to x, is - sin(x).

3) Using the "chain rule" to find the derivative of cos(3x- 1), you will have to multiply -sin(3x- 4) times the derivative of 3x- 4 which is 3.

HOI must have had a bit of a brain cramp. He meant to say "...to find the derivative of cos(3x - 4), you will..."

 

[Jason76]

\(\displaystyle f(x) = 2\cos(3x -4)\)

\(\displaystyle f'(x) = \dfrac{d}{dx}(2 \cos(3x - 4))\)

\(\displaystyle f'(x) = 2 [-\sin(u) du]\)

\(\displaystyle f'(x) = 2 [-\sin(u)(3)]\)

\(\displaystyle f'(x) = 6[-\sin(u)](3)\)

\(\displaystyle f'(x) = 6[-\sin(u)]\)

\(\displaystyle f'(x) = -6 \sin (u)\)

\(\displaystyle f'(x) = -6 \sin(3x - 4)\) Final Answer

 

[JeffM]

is the differentiation of 2cos(3x-4)=

1.6cos(3x-4)

2.-6sin(3x-4)


please help me and try explain it with step thanks

I shall try to explain it step by step in a way that I think is a bit clearer than Jason's.

\(\displaystyle y = 2cos(3x - 4).\)

Now if we let \(\displaystyle u = 3x - 4,\) we can substitute and get the simpler expression of \(\displaystyle y = 2cos(u).\)

\(\displaystyle y = 2cos(u) \implies \dfrac{dy}{du} = 2\{- sin(u)\} = -2sin(u).\) Right? Simple, correct?

But we want dy/dx, not dy/du. So we need the chain rule: \(\displaystyle \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx}.\)

We know what dy/du is. And we know that \(\displaystyle u = 3x - 4 \implies \dfrac{du}{dx} = 3.\)

\(\displaystyle So\ \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = -2sin(u) * 3 = - 6sin(u).\) But we want an answer in terms of x, not u.

\(\displaystyle THUS\ \dfrac{dy}{dx} = - 6sin(u) = - 6sin(3x - 4).\) Reverse the substitution.

Now when you get good at it, you can skip the intermediate steps and do the chain rule in your head.