How to Calculate Derivatives of Inverse Trigonometric Functions

How to Calculate Derivatives of Inverse Trigonometric Functions
Coming up next: Applying the Rules of Differentiation to Calculate Derivatives

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  • 0:06 Review of Basic Trig Functions
  • 1:57 Rules of Inverse Trig…
  • 2:37 Example #1
  • 4:25 Example #2
  • 5:51 Example #3
  • 7:05 Lesson Summary
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Lesson Transcript
Instructor: Zach Pino
Like a metronome, trigonometric functions are regular. Even predictable. In this lesson, you will learn how to use this predictability to remember the derivative formulas for these common functions.

Review of Basic Trig Functions

Three derivative formulas to remember.

I absolutely love trig functions, like sines and cosines. I love them because they repeat, they're very predictable and quite frankly, they're pretty graphs. But one of my biggest problems with trig functions is that it's really hard to solve the inverse.

Let's say I have the function y=cos(x), and I want to know, between zero and pi, what value of x is going to give me a y of 1/2. So I want to know what the value of x is right here. Well, to do that, I need to use the inverse trig function. So in this case, I've got x=cos(y) and this graph is the inverse of the previous one. I can write it as y=cos^-1(x). Now cos^-1 is not 1/cosine, but its the arccosine. It's a totally different function. That's great. Now I just look at x=0.5, and I can find what value of y is there and solve this problem.

Okay, so I've solved that problem, but what about taking the derivative of this function? I know that if I just have cos(x), I can find the derivative very easily. The derivative of cos(x) is just -sin(x), so that's the slope of the tangent here. But what about the slope of the tangent of the arccosine of x? So if y=cos^-1(x), what is y'?

Rules of Inverse Trig Functions

In example #1, simplify by multiplying out 4x^2 and moving the 4 on top of the fraction
Inverse Trig Derivative Example 1

Well here are three more derivative formulas for you to remember. For y=cos^-1(x), y`= -1 / the square root of (1 - x^2), as long as the absolute value of x is less than 1. If y=sin^-1(x), y`= 1 / the square root of (1 - x^2), again, as long as |x| < 1. For y=tan^-1(x), y`= 1 / (1 + x^2).

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