# Y = 2x arccos 3x Find the derivative

## Homework Statement

y = 2x arccos 3x Find the derivative. Can be left unsimplified.

## The Attempt at a Solution

y = 2x arccos 3x
y = 2x arccos (u) u = 3x
dy/du = d/dx(2x)arccos (u) + 2x[d/dx(arccos (u)] du/dx = 3
dy/du = 2arccos(u) + 2x[-1/sqrt of (1-u^2)]
dy/dx = 2arccos(3x) + 2x[-1/sqrt of (1-(3x)^2)](3)
dy/dx = 2arccos(3x) - [2x/sqrt of (1-(3x)^2)](3)
dy/dx = 6arccos(3x) - [2x/sqrt of (1-(3x)^2)]

## The Attempt at a Solution

e^(i Pi)+1=0
Try using the definition of the inverse to switch to a normal trig function, then use implicit differentiation. You'll be left with a sin(something) in your solution, which you can evaluate by drawing a triangle & finding out its value and subbing that back into your equation, then just isolate dy/dx and simplify.

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## Homework Statement

y = 2x arccos 3x Find the derivative. Can be left unsimplified.

## The Attempt at a Solution

y = 2x arccos 3x
y = 2x arccos (u) u = 3x
dy/du = d/dx(2x)arccos (u) + 2x[d/dx(arccos (u)] du/dx = 3
dy/du = 2arccos(u) + 2x[-1/sqrt of (1-u^2)]
dy/dx = 2arccos(3x) + 2x[-1/sqrt of (1-(3x)^2)](3)
dy/dx = 2arccos(3x) - [2x/sqrt of (1-(3x)^2)](3)
dy/dx = [STRIKE]6[/STRIKE] 2arccos(3x) - [[STRIKE]2[/STRIKE] 6x/sqrt of (1-(3x)^2)]
Some of your steps are strange, if not downright incorrect, [STRIKE]but you get to the correct result (with a corrected typo)[/STRIKE].

Edit: See the "strike out".

The first term should not be multiplied by 3. The second term should be.

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Staff Emeritus
Yes. there will be a sin(something). However, after simplifying the something you have sin(arccos(3x)), which is $\displaystyle \sqrt{1-9x^2\ }\ .$