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Math 104: Calculus13 chapters | 104 video lessons
Erin has taught math and science from grade school up to the post-graduate level. She holds a Ph.D. in Chemical Engineering.
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'?
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).
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.
Let's do another example. Let's say that f(x)=sin^-1(x^2). Well I see these parentheses, so again I'm already thinking chain rule. f`(x)=d/dx(sin^-1(x^2)). But I'm again going to use u instead of x^2, just to be clear here. So the derivative with respect to x of sin^-1(u) is 1 / the square root of (1 - u^2)du/dx - that's the chain rule. I have the derivative of the outside times the derivative of the inside. Now I'm going to plug in x^2 for u; I get 1 / the square root of (1 - x^2), that's u, all squared, times d/dx of x^2. This isn't so bad. Let's find this d/dx of x^2. I remember that that's just 2x, so I have 1 / the square root of (1 - (x^2))^2 * 2x. Well I can simplify this whole thing and find that f`(x)=2x / the square root of (1-x^4).
Let's do another example. Let's say this time I have f(x)=(tan^-1(x))^2. Okay, a lot of parentheses here - more chain rule. So f`(x)=d/dx(tan^-1(x))^2. I'm going to call tan^-1(x) u for a second. So I'm going to write this as d/dx of u^2. That looks much simpler. That's just 2u * du/dx. Okay, fantastic. Now I just plug in tan^-1 for u. So I have 2tan^-1(x) * d/dx * (tan^-1(x)). Well, let me use the rule I just learned for this d/dx of tan^-1. So I find that my f`(x) = 2tan^-1(x)(1/(x^2 + 1)). This simplifies to f`(x) = (2tan^-1(x))/(x^2 + 1).
So let's review. If y=sin^-1(x), that's like writing x=sin(y). We can find the derivative of this y=sin^-1(x) just by using this formula: y` (the derivative) = 1 / the square root of (1 - x^2). Similarly for cos^-1(x), the derivative is -1 / the square root of (1 - x^2). For tan^-1(x), it's 1 / (x^2 + 1).
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