AP®︎ Calculus BC (2017 edition)
Course: AP®︎ Calculus BC (2017 edition) > Unit 4Lesson 5: Derivatives of inverse trigonometric functions
Derivative of inverse cosine
Derivative of inverse cosine. Created by Sal Khan.
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- 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.(10 votes)
- 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.(70 votes)
- 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.(10 votes)
- 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.(20 votes)
- 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)?(5 votes)
- Well, if you have a negative function as
-sin(y), you could take
-1out 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.
* It will be clearer if you graph itm for example using Wolfram Alpha website which Sal mentioned*
To know dy/dx at any point we just substitute.
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.(8 votes)
- This confused me:
y = -arccos(x)
-y = arccos(x)
then the inverse is,
cos(-y) = x
due to symmetry:
cos(-y) = cos(y)
cos(y) = x
y = arccos(x)
-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?(3 votes)
- 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(10 votes)
- 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...(4 votes)
- The well established domain for inverse sin and cos is
-1 ≤ x ≤ 1.(6 votes)
- Why when you take
how does it become
Would that be to say d/dx[x]=(dx/dx)(d/dx[x])
because d/dx[cos(y)]=(dy/dx)(d/dx[cos(y)](3 votes)
- 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).(3 votes)
- isn't sin y is equal to + or - (sqrt 1-cos^2y)? then why Sal discard - (sqrt 1-cos^2y)?(4 votes)
- What is the derivative of inverse cosine of 1/x?? How can u find it??(2 votes)
- Observe that:
d/dx 1/x = -1/x²for
x ≠ 0, and
d/dx arccos(x) = -1/√(1 - x²)for
|x| < 1.
Now apply the chain rule.(4 votes)
- Hi this video was really helpful thanks but could someone please tell me how you would solve a similar problem if there was another x in front of the cos^-1(x)? E.g. if you were trying to solve y = x cos^-1(x) ? Thank you :)(2 votes)
- You now know the derivative of arccos(x), so this would just be an application of product rule.
As for solving y=x·arccos(x) for x algebraically, this cannot be done without introducing nonstandard functions or some other such thing.(4 votes)
- At2:00, Couldn't you have converted 1/sinY to cosY which is equal to X?(2 votes)
Voiceover: In the last video, we showed or we proved to ourselves that the derivative of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. What I encourage you to do in this video is to pause it and try to do the same type of proof for the derivative of the inverse cosine of x. So, our goal here is to figure out ... I want the derivative with respect to x of the inverse cosine, inverse cosine of x. What is this going to be equal to? So, assuming you've had a go at it, let's work through it. So, just like last time, we could write, let's just set y being equal to this. y is equal the inverse cosine of x, which means the same thing is saying that x is equal to the cosine, cosine of y. I'll just take the derivative of both sides with respect to x. On the left hand side, you're just going to have a 1. We're just going to have a 1. And on the right hand side, you're going to have the derivative of cosine y with respect to y, which is negative sine of y times the derivative of y with respect to x, which is dy, dx, and so we get ... Let's see if we divide both sides by negative sine of y, we get dy, dx is equal to negative 1 over sine of y. Now, like we've seen before, this is kind of satisfying, but we have our derivative in terms of y. We want it in terms of x. And we know that x is cosine of y, so let's see if we can rewrite this bottom expression in terms of cosine of y instead of sine y. Well, we know when we saw in the last video from the pythagorean identity that cosine squared of y plus sine squared of y is equal to 1. We know that sine of y is equal to the square root of 1 minus cosine squared of y. So, this is equal to negative 1. This is just a manipulation of the pythagorean trig identity. This is equal to 1 minus cosine. I can write like this, cosine squared of y, but I'll write it like this because it'll make it a little bit clearer. And what is cosine of y? Well, of course that is x, so this is equal to negative 1 over the square root of 1 minus. Instead of writing cosine y ... Instead of writing cosine y ... I'm trying to switch colors. Instead of writing cosine y, we could write 1 minus x, 1 minus x squared, so there you have it. The derivative with respect to x of the inverse cosine of x is ... I think I lost that color, I'll do it in magenta ... is equal to negative 1 over the square root of 1 minus x squared, so this is a neat thing. This right over here is a neat thing to know. And of course, we should compare it to the inverse, the derivative of the inverse sine. Actually, let me put them side by side, and we see that the only difference here is the sign, so let me copy and paste that. I'll copy and paste it, I'm going to paste it down here, and now let's look at them side by side. So, we see for taking the derivative with respect to x of the inverse cosine function, we have a negative. A negative 1 over the square root of 1 minus x squared. If we're looking at the derivative with respect to x of the inverse sine, it's the same expression except now it is positive.