Question

How do you differentiate `y=arcsin(x)`?

I don't even know where to start.

I watched a video on You Tube of how to do it, but it just didn't make any sense to me. Hope someone can help.

Thanks

Answer 1

See below.

Explanation:

`y=arcsin(x)`

Before we proceed we need to understand just what it is we are looking for. Remember that:

`y=arcsin(x)` is the inverse function of `y=sin(x)`

This can be expressed as:

`y=arcsin(x) <=> x=sin(y)`

Using

`x=sin(y)`

We need to differentiate in respect of `x`, so this will need to be differentiated implicitly.

Remembering that:

`d/dx(f(y))= (d)/(dy)(f(y))*dy/dx`

`dy/dx(x)=d/(dy)(sin(y))*dy/dx`

`1=cos(y)*dy/dx`

`dy/dx=1/cos(y)`

From above `y = arcsin(x)`

Substitutng in `dy/dx=1/cos(y)`

`:.`

`dy/dx=1/cos(arcsin(x))`

This is a little awkward, and it would be easier if we could express this in a different way.

Using the Pythagorean identity:

`color(red)(sin^2(y)+cos^2(y)=1)`

`cos(y)=+-sqrt(1-sin^2(y))`

Using positive root: ( see below)

`:.`

`dy/dx=1/(sqrt(1-sin^2(y)))`

From above `x=sin(y)`

Hence:

`dy/dx=1/(sqrt(1-x^2)`

`:.`

`dy/dx(arcsin(x))=1/(sqrt(1-x^2)`

Reason for using positive root of `sqrt(1-sin^2(y))`

This is because `y=sin(x)` only has an inverse if we restrict the domain to:

`-pi/2 <= x <= pi/2`

In `color(white)(88)1/cos(y)` , `color(white)(88)y` is an angle and it is an angle in the range

`-pi/2 <= x <= pi/2`

This range is in the I and IV quadrants, where the cosine ratios are positive.

Answer 2

Here's an alternative solution

Explanation:

This troubled me as well when I was first learning these kinds of derivatives. Here's how I learned how to tackle this type of problem:

Similarly to what was said below, `y=arcsin(x)` is just the inverse of `y=sin(x)`

When trying to take the derivative of `y=arcsin(x)` we can rewrite this as:

`color(blue)(sin(y)=x`

It is at this point where I would create a right triangle showing this

Recall: `sin(y)="Opposite"/"Hypothenuse"`

In this case `sin(y)=x/1`

We can solve for the missing side using `a^2+b^2=c^2` which will give you `sqrt(1-x^2)`. This will be useful later!

Now we'll differentiate implicitly (W.R.T `x`):

`dy/dx*cos(y)=1`

Divide `cos(y)` from both sides

`dy/dx=1/cos(y)`

Now we'll have to rewrite in terms of `x` but you may notice that we `color(red)(cos(y))` in the way but what is `color(red)(cos(y)`?

Well based on the triangle we made above:

`cos(y)="Adjacent"/"Hypothenuse"`

`color(red)(cos(y)=sqrt(1-x^2)/1=sqrt(1-x^2)`

So `dy/dx=1/cos(y)=1/sqrt(1-x^2)`

`:.dy/dx[arcsin(x)]=1/sqrt(1-x^2)`

The neat thing about this is that this is useful for the other inverse functions like `y=arccos(x)`

`y=arccos(x)<=>cos(y)=x`

Create a triangle:

Differentiate both sides implicitly (W.R.T `x`)

`dy/dx*-sin(y)=1`

Divide both sides by `-sin(y)`

`dy/dx=-1/sin(y)`

Rewrite in terms of `x`

Since `cos(y)=x/1`

Then `sin(y)=sqrt(1-x^2)/1=sqrt(1-x^2)`

So `dy/dx=-1/sin(y)=-1/sqrt(1-x^2)`

`:.dy/dx[arccos(x)]=-1/sqrt(1-x^2)`

Also if you have something like `y=arcsin(3x)` the same process applies but the triangle will be different based on whatever the coefficient is.

So...

`y=arcsin(3x)<=>sin(y)=3x`

Differentiate implicitly (W.R.T `x`):

`dy/dx*cos(y)=3`

`dy/dx=3/cos(y)`

Since `sin(y)=(3x)/1`

Then `cos(y)=sqrt(1-9x^2)/1=sqrt(1-9x^2)`

So `dy/dx=3/cos(y)=3/sqrt(1-9x^2)`

`:.dy/dx[arcsin(3x)]=3/sqrt(1-9x^2)`

I hope I explained this well!

Answer 3

`dy/dx = 1/cos(arcsinx)=1/sqrt(1-x^2)dx`

Explanation:

There is another way to do it. We have a formula for calculating the derivatives of inverse functions:

`(f^-1)^'(x)=1/(f^'(f^-1(x))`

Here `f^-1(x)=arcsinx rArr f(x)=sinx` and `f^'(x)=cosx`

So `(arcsinx)^'=1/cos(arcsinx)`

Using the arguments given in the other answers, we know that

`1/cos (arcsinx)=1/sqrt(1-x^2)`