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How to Calculate Derivatives of Inverse Trigonometric Functions

  • 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: Erin Lennon

Erin has taught math and science from grade school up to the post-graduate level. She holds a Ph.D. in Chemical Engineering.

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).

Example #1

Let's do an example. Let's say that f(x)=cos^-1(4x). So f`(x) is the derivative, d/dx, of cos^-1(4x). Now this example is a little bit trickier than it lets on at first. I'm going to call 4x u for a second, because I know I've got parentheses here and I'm going to have to use the chain rule. I'm going to write this as d/dx of cos^-1(u). Using the chain rule, I'm first going to find the derivative of cos^-1(u). That's -1 / the square root of (1 - u^2) because that's the rule we just learned. I'm going to multiply all that by the derivative of what's inside of the parentheses using the chain rule; that's du/dx. Okay, so I've gotten rid of my parentheses. Let's plug 4x back in for u, so we get -1 / the square root of (1 - 4x^2), all times d/dx of 4x. Well d/dx of 4x is just 4, and I can simplify this whole thing by multiplying out this 4x^2 and moving the 4 over on top of this fraction. I get f`(x)=-4 / the square root of (1 - 16x^2). Hmm. Not exactly what I was expecting, but it's all because I used the chain rule here.

The process of solving example #2
Inverse Trig Derivative Example 2

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