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### Course: Differential Calculus>Unit 3

Lesson 6: Derivatives of inverse trigonometric functions

# Derivative of inverse cosine

Unraveling the mystery of the inverse cosine function, we find its derivative equals -1/(sqrt(1 - x^2)). This step-by-step proof guides us to a fascinating comparison with the derivative of inverse sine, revealing a captivating connection between these two trigonometric functions. Created by Sal Khan.

## Want to join the conversation?

• Thanks for the tutorial but i was wandering why couldn't we just differentiate arccos x directly by changing it from arccos X to 1/cosX and then use quotient rule to get the derivative.
• No, because cos⁻¹ x is NOT the same as (cos x)⁻¹. Therefore,
arccos x is NOT 1/cos x.

To be precise,
arccos x = 𝑖 ln [ x - 𝑖√(1-x²) ]

The ⁻¹ beside the name of a function refers to the inverse function, not the reciprocal. So, this notion is NOT an exponent. This inconsistency in notation often causes confusion with students.
• Sal again did not specify the reason why we could just take the principal root of
1 - cos^2(y). Since the slope of the inverse cosine function within its restricted range without the endpoints is always negative, we have to take the positive root, so that the fraction ends up negative, but we should at least consider the possibilty of using the negative root, and then exclude it because the result would be positive.
• I agree Sal should have included the steps explaining why we only took the principal root and not the negative (I don't remember if he did on the inverse sine video). Here is why.

Remember back in the inverse functions chapter, when you invert a function, its domain and range switch places.

y = arccos(x)
When we convert this into x = cos(y), we have to add a restriction (0 ≤ y ≤ π) - my college instructor is nitpicky and would take point off if you don't add this restriction in the proof.

cos²(y) + sin²(y) = 1 (0 ≤ y ≤ π)
Because of that restriction, sin(y) is always positive (top half of the unit circle from 0 ≤ y ≤ π ) so you ignore the negative root.

Hope this helps.
• What if you left the -sine(y) in the initial derivative in the denominator and tried to get your answer in terms of X? Does this method work for negative trigonometric functions like -sine(y)?
• Well, if you have a negative function as `-sin(y)`, you could take `-1` out of a derivative, as it is a constant, so you get
`dy/dx(-1sin(y))= -1 dy/dx(sin(y))= -1 * cos(y)= -cos(y)`

As for the first part of you question (as far as I understood it), you had to see the sin(y) in terms of X so you will be able to tell the actual value of a derivative for any X. Otherwise, if you just leave there `1/-sin(y)` to get a right result of the derivative you need to put that Y, that correspond to the point of the graph the derivative of which you are trying to find. However, this raises a question if there are two points on the graph for the same Y.
As if `y=x^2` (parabola):
* It will be clearer if you graph itm for example using Wolfram Alpha website which Sal mentioned*
y=x^2
x=sqrt(y)
dy/dx(x^2)=2x
so 2x=2sqrt(y)

To know dy/dx at any point we just substitute.
For example,
X: dy/dx at (0.5 , 0.25) = 2 * 0.5=1
Y: dy/dx = 2 * sqrt(0.25) = 1

It seems OK, but remember: this is Parabola, so we have 2 points at Y = 0.25. And the derivative of one is (1), the derivative of other (-1) so we have 2 X for each Y.
Thus it is not a problem if we use X, as there is Only one Y for each X.

Hope I did everything right.
• This confused me:

`y = -arccos(x)`
then,
`-y = arccos(x)`
then the inverse is,
`cos(-y) = x`

due to symmetry:
`cos(-y) = cos(y)`
thus,
`cos(y) = x`
and,
`y = arccos(x)`

BUT:
`-arccos(x) != arccos(x)`

Am I just missing some property of inverse trigonometric functions that doesn't allow some part of what I did to occur?
• The problem lies in the domain of the cosine function.
In order for cos to be invertible we have to restrict its domain to [0,π]. Without this restriction arccos would be multivalued.
Because of this restriction your "due to symmetry: cos(-y) = cos(y)" assertion is no longer true, since either y or -y must be outside that domain.

What you've done is a bit like saying √x = -√x because (√x)² = (-√x)²
• What I did was convert -sin(x) to sqrt(x^2-1) on the left hand side itself and then divided both sides by it... That way I got 1/sqrt(x^2-1) ... Now this SHOULD be similar to -1/(sqrt(1-x^2) but I don't know why, all I know is the algebra works out... (If it doesn't, please tell me why). My only concern is that if they are to be equal, the sqrt should return a negative term (that is, if sqrt(1-x^2) is positive then sqrt(x^2-1) should be negative but wouldn't that require the use of i?
I don't know if I've just done something stupid that mathematicians usually just avoid or done a sensible step but can't justify it...
• The well established domain for inverse sin and cos is `-1 ≤ x ≤ 1`.
• Why when you take
d/dx[cos(y)]=d/dx[x]
how does it become
(dy/dx)(-sin(y))=1
Would that be to say d/dx[x]=(dx/dx)(d/dx[x])
because d/dx[cos(y)]=(dy/dx)(d/dx[cos(y)]
• Yes, that is exactly correct. That's the chain rule. We never actually write the dx/dx, since we can see it will be 1, and we only need the chain rule when there is a "mismatch", as when d/dx is applied to cos(y).
• isn't sin y is equal to + or - (sqrt 1-cos^2y)? then why Sal discard - (sqrt 1-cos^2y)?
• is it a negative one over the square root of 1-x^2. I know that is what he said in the video but the way he drew it made me think it was the entire fraction that was negative. Just need to clarify thanks!
• You know how negative numbers work right?

It does not matter although it might slightly simpler the negative sign outside the numerator.

i.e. Instead of (-1)/2 it is sligthly better to write -(1/2)
• What is the derivative of inverse cosine of 1/x?? How can u find it??
• Observe that:
`d/dx 1/x = -1/x²` for `x ≠ 0`, and
`d/dx arccos(x) = -1/√(1 - x²)` for `|x| < 1`.
Now apply the chain rule.