We just learned our rules for finding the derivatives of inverse sine and inverse cosine, and we found that we could actually find what those rules are using the process of implicit differentiation. And we know that there are other inverse trig functions besides sine and cosine. We have the inverse tangent, inverse cotangent, inverse secant, and inverse cosecant. And we could follow this same exact process of implicit differentiation in order to find a derivative rule for all of these other inverse trig functions. And you can feel free to go through that process on your own.
Here, we're going to work through these remaining derivatives for our inverse trig functions in practice, alongside all of the other rules for derivatives that we've already learned so far. So let's go ahead and jump right into our first example. Here, we want to find the derivative of \( f(x) = 4x^2 \cdot \text{arctan}(x) \). Now here, I know since I have 2 functions being multiplied, that I need to use the product rule in order to find my derivative \( f'(x) \). Remember, our memory tool for the product rule tells us left d right plus right d left.
So I want to take that left-hand function \( 4x^2 \) and multiply it by the derivative of my right-hand function, that inverse tangent of \( x \). Now the derivative of the inverse tangent of \( x \) is going to be \( \frac{1}{1 + x^2} \). So I'm multiplying this \( 4x^2 \) by \( \frac{1}{1 + x^2} \) using that rule for the inverse tangent. Then continuing on with my product rule here, I have a plus, a right d left, so that's the inverse tangent of \( x \), times the derivative of this \( 4x^2 \), which is \( 8x \). Now I can rewrite this a little bit more compactly with that \( \frac{4x^2}{1 + x^2} \), and then adding this together, moving that \( 8x \) to the front, \( 8x \cdot \text{arctan}(x) \).
And this is my full derivative here, \( f'(x) \), having used my product rule alongside my new rule for the inverse tangent. Let's work through another example and take a look at some other inverse trig functions. Here, we want to find the derivative of \( g(x) = 6 \cdot \text{arccot}(x) + 4 \cdot \text{arcsec}(3x + 1) \). Now here, since I have 2 derivatives being added together, that means that I need to add their derivatives.
So \( g'(x) \) here is going to be equal to 6 times the derivative of the inverse cotangent. Now looking back up to my table here, the derivative of the inverse cotangent is \(-\frac{1}{1 + x^2}\). So it's just the negative of what the derivative of inverse tangent was. So this gives me \( 6 \cdot \left(-\frac{1}{1 + x^2}\right) \). Then adding this together with this second derivative here, the derivative of 4 times the inverse secant of \( 3x + 1 \), pulling that constant out front, then I look for the inverse secant coming up to my table here.
The derivative of the inverse secant is going to be \( \frac{1}{|\text{interior function}| \cdot \sqrt{\text{interior function}^2 - 1}} \). Now here, since I have the inverse secant of \( 3x+1 \) and not just \( x \), I need to keep that in mind as I take this derivative. So this gives me \( \frac{1}{|3x + 1|} \times \text{derivative of } (3x + 1)^2 - 1 \). I need to apply the chain rule here, working to that inside function, multiplying by the derivative of \( 3x + 1 \), which is just 3.
Now we can clean this up a little bit. In this first term, 6 times -1 is -6. So this is \( -\frac{6}{1 + x^2} \). Then for this second term here, multiplying 4 times 3 gives me 12 in my numerator.
My denominator is rather long. It is the absolute value of \( 3x+1 \) times the square root of \( (3x + 1)^2 - 1 \). And that's my full derivative here, \( g'(x) \). Now we have one final example to work through. We want to find the derivative of this function \( h(x) \), which is the inverse cosecant of \( x \), and all of that is getting cubed.
Now in order to find our derivative here, since this stuff is being cubed, we need to start with our power rule. So to find our derivative here, \( h'(x) \), we're going to pull that 3 out front using the power rule, and then decrease it by 1. This getKeyframeAreaBoxSize \( 3 \cdot (\text{arccsc}(x))^2 \), having used that power rule. Then using the chain rule, I need to multiply by the derivative of the inverse cosecant of \( x \). This is our last remaining rule for our inverse trig functions.
Now the derivative of the inverse cosecant is going to be \(-\frac{1}{|x| \cdot \sqrt{x^2 - 1}}\). So again, here, we can see that it's just the negative version of the derivative of the inverse secant. So multiplying by this, this is \(-\frac{1}{|x| \cdot \sqrt{x^2 - 1}}\). That inside function is just \( x \). So I can just use that rule as is.
Now I can rewrite this altogether in one fraction. This gives me \(-3 \cdot (\text{arccsc}(x))^2 \over |x| \cdot \sqrt{x^2 - 1}\). And this is my final derivative for this function, \( h'(x) \). Now, I have one more thing to mention here. Just as we had a restriction on our \( x \) value for the inverse sine and inverse cosine, we also have one for the inverse secant and inverse cosecant.
For these two derivatives, the absolute value of \( x \) must be greater than 1 rather than less than 1. And this is based on the original domains of these functions. Now as I've said before, whenever we're working with derivatives of trig functions, it's going to be easiest to memorize them so that we can work through examples quicker. But remember that if you're ever unsure of what these derivatives are, you can go back through your process of implicit differentiation. Either way, this process is going to get easier and quicker as we work through additional examples, and that's exactly what we're doing coming up next.
I'll see you there.
