Taking the Derivative of arcsin: How-To & Tutorial

Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

In this lesson, the derivative of arcsin x is determined using a reference triangle and the chain rule. These ideas are extended by plotting arcsin x and then using the derivative result to draw the tangent line.

Derivative of Arcsin

The steps for taking the derivative of arcsin x:

Step 1: Write sin y = x,


sin_y_=_x


This might look strange. We are used to writing y is equal to some function of x like y = sin x. Instead, we are writing some function of y is equal to x. The reason we do this is because arcsin x is not easy to work with. But we know that arcsine and sine cancel each other out, and sine is a lot easier to work with, so we just multiply both sides of the equation by sine to get sin y = x.

Step 2: Differentiate both sides of this equation with respect to x.


derivative_of_sin_y_=_x


On the left-hand side, using the chain rule, the derivative of sin y is cos y times dy/dx. On the right-hand side, the derivative of x is 1.


cos_y_dy/dx_=_1


Step 3: Solve for dy/dx.

Make dy/dx the subject by dividing both sides by cos y:


dy/dx_=_1/cos_y


Step 4: Define cos y in terms of x using a reference triangle.

A reference triangle shows the angle y. The sine of y is the opposite side over the hypotenuse. Thus, the reference triangle has a hypotenuse of 1 and an opposite side of x. The adjacent side is found using Pythagorus' theorem. The completed reference triangle:


The reference triangle
the_reference_triangle


The cosine of y is the adjacent side divided by the hypotenuse. From the reference triangle, the adjacent side is √(1 - x2) and the hypotenuse is 1. Thus, cos y = √(1 - x2)/1 which means 1/cos y = 1/√(1 - x2):

Step 5: Substitute for cos y.


dy/dx_=_1/square_root(1-x^2)


Step 6: Define arcsine.

Now we can go back to our arcsin x, which can also be written as the ''angle whose sine is x'' or sin-1.

Using this notation, y = sin-1x. Exchanging left- and right-hand sides:


arcsin_x_=_y


Step 7: Differentiate and write the result.

Differentiate both sides with respect to x:


differentiate_arcsin_x_=_y


But we've just found dy/dx equals 1/√(1 - x2):


derivative_arcsin_x_=_1/(1-x^2)^.5


This is a good result but we need some additional ideas to make the answer complete.

What about the allowed values for x? Clearly, the derivative of arcsin x must avoid dividing by 0: x ≠ 1 and x ≠ -1. But also, because sin x is bounded between ± 1, we won't allow values for x > 1 nor for x < -1 when we evaluate arcsin x.

This may be clearer if we plot out arcsin x, or, as we learned above, sin y = x.

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