Question 1

At 3:05 : What assures us that it shouldn't be plus and minus the square root? Like this: cos(y)=+-sqrt(1-sin(y)^2)?

Accepted Answer

Good question! Sal skipped over the reason, and it's far from obvious.

The answer has to do with the way we define the inverse sine function. As you know, the sine function is cyclical, so we wouldn't get a valid function if we created its inverse in the usual way of simply reflecting it across the y = x diagonal because we wouldn't have a unique output for valid inputs. For example, the inverse sine of 0 could be 0, or π, or 2π, or any other integer multiplied by π. To solve this problem, we restrict the range of the inverse sine function, from -π/2 to π/2. Within this range, the slope of the tangent is always positive (except at the endpoints, where it is undefined). Therefore, the derivative of the inverse sine function can't be negative. We can safely discard the negative square root in this derivation because it would give us a negative derivative, which is impossible because of the way we've restricted the domain of the inverse sine function.

Question 2

I am a bit rusty with taking derivatives of inverse functions but is there a reason why when taking the derivative of y=(sin x)^-1 that we could not use the Power or Quotient Rule to solve the function?  When I use the power/quotient rule I get d(y)/dx = d((sin x)^-1)/dx = -cos x/(sin x)^2 and not 1/sqrt(1-x^2) that Sal mentions.  Where am I going wrong?

Accepted Answer

The inverse functions, though written as sin⁻¹, etc. ARE NOT the reciprocals of those functions. They are NOT being raised to the -1 power. Thus, what you were doing was finding the derivatives of the reciprocal functions, not the inverse functions.

So, remember that sin⁻¹ x is NOT (sin x)⁻¹ and is NOT 1 / sin x.
To avoid confusion, you can use the alternative notation of arc-
sin⁻¹ x = arcsin x
 
The same goes for all of the other trig functions.

Question 3

Hi everyone, maybe I'am asking this question because I just had not watched the videos about trigonometry but maybe someone can give me a quick explanation. So, why y=sin^(-1)x is the same as siny=x

Accepted Answer

Suppose _y=x²_ . This is y in terms of x.
Now if you want to find out what x is in terms of y, then solve for x to get _x=√y_.
As you know, the square operator and the square root operator are inverses of each other, that is, one "undoes" the other: _√(x²) = (√x)² = x_ (assuming we are only interested in the _principal_ square root).

It is the same deal with _sin_ and _arcsin_, which is conventionally written as _sin^-1 x_.
_Arcsin_ is the inverse of _sin_, such that _arcsin(sin(x)) = x_, or _sin(arcsin(x))=x_.
Like the square/square root example, if you have _y=sin(x)_, which is y in terms of x, but you want to take that expression and find x in terms of y, then given:

_y=sin(x)_
take the arcsin of both sides:
_sin^-1(y)=sin^-1(sin(x))_, so that: _sin^-1(y)=x_

or, using the term _arcsin_ and not _sin^-1_ (though _sin^-1_ is more common)
_y=sin(x)_
_arcsin(y) = arcsin(sin(x)) = x_, so that
_arcsin(y) = x_

It is important to know the inverse trig functions as they come in handy in many situations, like trig substitution in integral calculus.

Question 4

At 2:40, you begin to explain that you want to change dy / dx = 1 / cos y into something in terms of x by using the Pythagorean Identity and setting that (the Pythagorean Identity) equal to cos y . Could you just have inserted your original definition of y (y = sin^-1 x) instead? You would end up with:

dy / dx = 1 / (cos (sin^-1 x)).

Does that work ?  tried it with a few arbitrary numbers for x and was getting the same answer for both equations, but wasn't sure if it technically was correct.

Accepted Answer

Yes. If you want to visualize, try to type `cos(arcsin(x))` and `sqrt(1-x^2)` into Google or `is cos(arcsin(x)) equal to sqrt(1-x^2)` into WolframAlpha.

Question 5

When prompted in the beginning, I solved dy/dx for (sinx)^-1, and got a different result.
I looked into it and realized that (sinx)^-1 and sin^-1(x) are different things.

How come is (sinx)^-1 =/= sin^-1(x) but (sinx)^2 = sin^2(x) ?
Did someone deliberately come up with notations that are as confusing as possible just to mess with me or am I missing something?

Accepted Answer

Inverse functions are, perhaps unfortunately, often denoted with the superscript `-1`. This can be confusing at times if it is not made clear whether we mean to denote the inverse function or the reciprocal function. However, the inverse trigonometric functions, known as the _arcus_ functions, are typically denoted `arcsin`, `arccos`, etc. So instead of writing the inverse of the sine function at `x` as `sin^(-1) x`, one can also write `arcsin x`, which is better practice, I would argue.

Question 6

I'm a little rusty on inverse trig functions, but why is it that he can write y=sin^(-1)x as sin(y)=x?

Accepted Answer

First and foremost, know that the inverse sine function is not the sine function to the negative 1 power. It's better if you write out inverse sine rather than sine to the negative one power. But to answer your question, if y=arcsin(x) which means the same thing as inverse sine, then sin(y)=x. That was the first step of the problem.

Question 7

Do mathematician needs cheat sheet while solving calculus?

Accepted Answer

Derivatives? No, everyone generally does it from memory. Integral tables? Yes. No one has the time to memorize thousands of different forms, even if they are less commonly used.

Question 8

Is there any video on Khan Academy about the derivatives and integrals of hyperbolic functions? Another question: Can complex numbers be differentiated or integrated?

Accepted Answer

The hyperbolic functions are usually defined in terms of the exponential function. As such, their properties are easily deduced from properties of the exponential function.

Complex-valued functions defined on a subset of complex numbers may (some times) be differentiated, yes. Let C denote the set of complex numbers, and suppose U is some subset of C. Suppose ƒ: U → C is a complex-valued function defined on U, and suppose w is an interior point of U. If the limit

lim (z → w) [ƒ(z) - ƒ(w)] / [z - w]

exists and is finite, we say that ƒ is differentiable at w, and we denote the value of this limit by ƒ'(w). If U is open, and if ƒ is differentiable on all of U, we say that ƒ is differentiable (on U). If ƒ is differentiable on an open set U, one also says that ƒ is holomorphic on U, or sometimes that ƒ is analytic on U. Holomorphic functions are central in the theory of complex functions.

More specifically, to say that ƒ: U → C is differentiable at an interior point w in U means the following: there exists some complex number L such that for every real number ε > 0 there exists a real number δ > 0 with the property that for all complex numbers z in U with 0 < |z - w| < δ, we have |[ƒ(z) - ƒ(w)]/[z - w] - L| < ε. If such a number L exists, we usually denote it by ƒ'(w). This property may also be cast in terms of convergent sequences in U.

The process of differentiation of complex-valued functions defined on subsets of the complex numbers share many properties with differentiation of real-valued functions defined on subsets of the real numbers. For instance, the differentiation operator is linear. Furthermore, the product rule, the quotient rule, and the chain rule all hold for such complex functions.

I will not include a discussion on integration of complex-valued functions defined on subsets of C, as this would require more sophisticated typesetting than what is available here.

Question 9

Once one gets to dy/dx = cos^-1(y), could the original identity y=sin^-1(x) be substituted back in? So it would be dy/dx = cos^-1(sin^-1(x))? Is this equivalent? Thanks.

Accepted Answer

Your idea is correct, but your final answer is not.
Just to remember: 
`cos⁻¹(x)` is NOT equal to `(cos(x))⁻¹`.
`cos⁻¹(x)` is equal to `arccos(x)`
The correct answer would be: `dy/dx = cos(sin⁻¹(x))⁻¹`

Question 10

Is there anything like an inverse derivative? For example, the inverse derivative of cos(x) would be sin(x)? Thank you. Please also provide a detailed explanation. Again, thanks in advance!

Accepted Answer

Just like addition and subtraction are inverse functions, multiplication and division are inverse functions, and derivatives and integrals are inverse functions. Take a look at integral calculus to learn more about integrals. :) It's not quite as cut-and-dry as the other inverse operations (think about the derivative of f(x)=x^2+5 being f'(x)=2x, and you realize that going backward you wouldn't know there would be a '+5').

Differential Calculus

Course: Differential Calculus > Unit 3

Derivative of inverse sine

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