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Calculus: Help and Review13 chapters | 148 lessons

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.

The steps for taking the derivative of arcsin *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*.

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

Make d*y*/d*x* the subject by dividing both sides by cos *y*:

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 cosine of *y* is the adjacent side divided by the hypotenuse. From the reference triangle, the adjacent side is √(1 - *x*2) and the hypotenuse is 1. Thus, cos *y* = √(1 - *x*2)/1 which means 1/cos *y* = 1/√(1 - *x*2):

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-1*x*. Exchanging left- and right-hand sides:

Differentiate both sides with respect to *x*:

But we've just found d*y*/d*x* equals 1/√(1 - *x*2):

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*.

Keep in mind *y* is the angle in radians.

- For
*y*= 0, sin*y*= sin 0 = 0 so*x*= 0 for*y*= 0 - For
*y*= π/2 radians, sin π/2 = 1 so*x*= 1 for*y*= π/2 radians - For
*y*= -π/2 radians,*x*= sin (-π/2) = -1 for*x*= -1 for*y*= -π/2 radians

*y* goes from -∞ to +∞ but *x* never gets larger than +1 nor smaller than -1.

Note: |*x*| < 1 means the same as -1 < *x* < 1.

You can use the derivative of arcsin *x* to plot the line tangent to arcsin *x* at any point. Let's pick *x* = 1/2.

- At
*x*= 1/2,*y*= arcsin (1/2) = .5236. - The slope
*m*at*x*= 1/2 is*m*= 1/√(1 - .52) = 1.1547. - Solving
*y*= arcsin*x*and*y*=*m**x*+*b*at*x*= 1/2 and*y*= .5236 gives*b*= -.0538.

Thus, the tangent line at *x* = 1/2 is *y* = 1.1547*x* - .0538.

The slope will always be positive because we take the positive square root. Note at *x* = ±1, the slope = +∞. This agrees with the our result for the derivative of arcsin *x* evaluated at these two points.

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Calculus: Help and Review13 chapters | 148 lessons

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