## d/dx arcsin(x)

### To find the derivative of the function `arcsin(x)` with respect to `x`, we can use the chain rule

To find the derivative of the function `arcsin(x)` with respect to `x`, we can use the chain rule.

First, let’s define the function `y = arcsin(x)`. The arcsin function returns the angle whose sine is `x` (in radians) and is defined for values between -1 and 1.

To find the derivative, we need to differentiate `y` with respect to `x`. We can rewrite the arcsin function in terms of sine:

sin(y) = x

Applying the derivative to both sides of the equation with respect to `x` gives:

d/dx(sin(y)) = d/dx(x)

Now, let’s apply the chain rule. The chain rule states that if `y = f(g(x))`, then the derivative of `y` with respect to `x` is given by:

dy/dx = f'(g(x)) * g'(x)

In our case, `f(u)` is the sine function and `g(x)` is the arcsin function. Therefore, `g'(x)` is the derivative of `arcsin(x)`.

Using the chain rule, we have:

cos(y) * dy/dx = 1

Now, we need to solve for `dy/dx` to find the derivative of `arcsin(x)`. Rearranging the equation, we get:

dy/dx = 1 / cos(y)

To find the value of `cos(y)`, we need to consider the trigonometric identity:

sin^2(y) + cos^2(y) = 1

Since we know that `sin(y) = x` from the original equation, we can substitute this into the identity:

x^2 + cos^2(y) = 1

Rearranging the equation, we get:

cos^2(y) = 1 – x^2

Taking the square root of both sides gives:

cos(y) = sqrt(1 – x^2)

Substituting `cos(y)` back into the derivative equation, we have:

dy/dx = 1 / sqrt(1 – x^2)

And this is the derivative of `arcsin(x)` with respect to `x`.

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