If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: AP®︎/College Calculus BC>Unit 3

Lesson 5: Differentiating inverse trigonometric functions

# Differentiating inverse trig functions review

Review the derivatives of the inverse trigonometric functions: arcsin(x), arccos(x), and arctan(x).

## What are the derivatives of the inverse trigonometric functions?

$\frac{d}{dx}\mathrm{arcsin}\left(x\right)=\frac{1}{\sqrt{1-{x}^{2}}}$
$\frac{d}{dx}\mathrm{arccos}\left(x\right)=-\frac{1}{\sqrt{1-{x}^{2}}}$
$\frac{d}{dx}\mathrm{arctan}\left(x\right)=\frac{1}{1+{x}^{2}}$

## Want to join the conversation?

• Why were arccsc(x) & arcsec(x) left out? Often ignored, but they're apart of the trig family too!
• Probably because it's actually really confusing. Think about it: Take arcsec(x). d/dx (1/cos(x)) would be a quotient of derivatives. I presume you know the complicated equation for that. Stuff arcsec(x) into it. Yeah. Also you'd probably rarely see it on the AP test.
• What about the reciprocal trig inverses? Can you provide videos/text about arccot(x), for example?
• d/dx arccot(x) = - 1 / (1+x²)

d/dx arcsec(x) = 1 / (x√(x²-1)) ; for 0≤x<π/2 and π≤x<3π/2
d/dx arcsec(x) = 1 / (|x|√(x²-1)) ; for 0≤x<π/2 and π/2<x≤π

d/dx arccsc(x) = - 1 / (x√(x²-1)) ; for 0≤x<π/2 and π≤x<3π/2
d/dx arccsc(x) = - 1 / (|x|√(x²-1)) ; for 0≤x<π/2 and π≤x<3π/2

We often use the first case in college however. The quadrants determine tan function positive or negative in the differentiation. The first restriction is QI and QIII, so tan is always positive, thus we have x without the absolute value before the radical. The second restriction is QI and QII, tan can either be positive or negative, thus we have |x|.

Another thing to remember that the derivatives of the "co-" arc-trig functions is just the negative of their counterparts. See how the derivative of arccos(x) is just negative of what arcsin(x) has, similar for arctan(x) and arccot(x), and arcsec(x) and arccsc(x)
• ngl after a bit of practice these have to be some of the easier things to do
• Why do we call inverse trig functions as arctrig functions?
• You see regular trig functions represent a ratio. Arctrig functions represent an angle. In a way, an arc is an angle which has been given an extra dimension of radius. If you were to zoom out while looking at an arc it will look like an angle.
• could you give an example on how to solve more difficult questions? for example find the derivative of : arcsin(x) / arcsin (2x).
• how can this be applied in real life?
• Maybe someones heartbeat can be represented by a trigonometric function, and you want need to report to a doctor the rate at which the patient's heart rate is increasing at a moment in time, so the doctor can perform his procedure when the rate is calm and steady. (I don't know the exact job of a doctor, but something like that)
• I am wondering, if not going straight for the rules above, how I can figure out the derivative of arcsin(-3x). I was hoping to use the trig rule and chain rule but then I got stuck...
• f(x) = arcsin(u) and u = -3x
Using the arcsin trig rule and chain rule:
f'(x) = d/dx (arcsin(-3x)) * du/dx = (1/√(1-(-3x)²)) * -3 = -3/√(1-9x²)
• What are the derivative for arccot, arcsec(x), and arccsc(x)?
• Here are how the rest of the inverse trig functions are differentiated. Make sure you have something to read LaTeX:

$\large\color{green}{\textbf{arccot(x)}}$

Let $y=arccot(x)$. So, $cot(y) = x$. DIfferentiating both sides w.r.t x, we get

$\frac{d}{dx}cot(y) = 1$

Using the chain rule, we have

$-cosec^{2}(y)\frac{dy}{dx}$ = 1

$\frac{dy}{dx}$ = $\frac{-1}{cosec^{2}(y)}$

Now, y was something we introduced, so we need to remove it. See that $cot(y) = x$. Drawing a triangle satisfying this, we get the follow results:

adjacent side = $y$

opposite side = $1$

hypotenuse = $\sqrt{1+x^{2}}$

Now, cosec(x) is defined to be $\frac{hypotenuse}{opposite}$. So, $cosec(x) = \sqrt{1+x^{2}}$ and $cosec^{2}(y) = 1+x^{2}$

Now, we already derived that $\frac{dy}{dx}$ = $\frac{-1}{cosec^{2}(y)}$. Substituting the value of $cosec^{2}(y)$ that we got, we have

$\boxed{\mathbf{\frac{d}{dx}{arccot(x)} = \frac{-1}{1+x^{2}}}}$

$\large\color{green}{\textbf{arcsec(x)}}$

Really similar logic. Let $y=arcsec(x)$. So, $sec(y) = x$. Differentiating both sides w.r.t x, we get

$\frac{d}{dx}sec(y) = 1$

$sec(y) \cdot tan(y) \frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{1}{sec(y)tan(y)}$

Again, y is something we need to remove. So, drawing a triangle with $sec(y) = x$, we have the following:

opposite = $\sqrt{x^{2}-1}$

hypotenuse = x

Now, we need to find sec(y) and tan(y). As we know their definitions, we have $sec(y) = x$ and $tan(y) = \sqrt{x^{2}-1}$. Substituting this into $\frac{dy}{dx} = \frac{1}{sec(y)tan(y)}$, we get,

$\frac{dy}{dx} = \frac{1}{x\sqrt{x^{2}-1}}$

Now, we would be done, but there's a small issue. arcsec(x) is defined for negative numbers too and the slope of the graph for negative numbers is positive. Essentially, even for $x<0, \frac{dy}{dx}>0$. How do we fix this? Well, by putting an absolute value sign on the "x" in the denominator. Now, the x under the square root can never be negative (as it is being squared). So, the x outside the square root dictates the sign of the derivative. So, that's what gets the absolute value. This gives us the derivative of arcsec(x) as:

$\boxed{\mathbf{\frac{d}{dx}arcsec(x)=\frac{1}{\mid x \mid\sqrt{x^{2}-1}}}}$

$\large\color{green}{\textbf{arccosec(x)}}$

Rinse and repeat. Take $y=arccosec(x)$ and hence, $cosec(y) = x$. Differentiating w.r.t x, we get:

$\frac{d}{dx}cosec(y) = 1$

$\frac{-cosec(y)}{cot(y)} \frac{dy}{dx} = 1$

$\frac{dy}{dx} = \frac{-1}{cosec(y)cot(y)}$

As usual, we need to remove y. So, draw a triangle satisfying $cosec(y) = x$ and we have the following observations:

adjacent = $\sqrt{x^{2}-1}$

opposite = 1

hypotenuse = x

Now, we need both $cosec(y)$ and $cot(y)$. From the data above and the definitions of the trigonometric ratios, we have $cosec(y) = x$ and $cot(y) = \sqrt{x^{2}-1}$. Substituting this into $\frac{dy}{dx} = \frac{-1}{cosec(y)cot(y)}$, we get,

$\frac{dy}{dx} = \frac{-1}{x\sqrt{x^{2}-1}}$

Again, we run into the same problem as arcsec(x). Hence, we add a modulus sign to the x to give us our final answer:

$\boxed{\mathbf{\frac{d}{dx}arccosec(x) = \frac{-1}{\mid x \mid\sqrt{x^{2}-1}}}}$
• I can get all the answers correct now. But How long will it take before I forget all about it?
• I am assuming that you are asking about remembering formulas for differentiating inverse trig functions.

If you forget one or more of these formulas, you can recover them by using implicit differentiation on the corresponding trig functions.

Example: suppose you forget the derivative of arctan(x). Then you could do the following:

y = arctan(x)
x = tan(y)
1 = sec^2(y) * dy/dx
dy/dx = 1/sec^2(y)
dy/dx = 1/[tan^2(y) + 1]
dy/dx = 1/(x^2 + 1).

So the derivative of arctan(x) is 1/(x^2 + 1).
• please prove the case when x>0 , y<0 and xy<-1

then: arctan(x) - arctan(y) = pi + arctan[(x-y)/(1+xy)]
• arctan(x) - arctan(y) = pi + arctan[(x-y)/(1+xy)]
tan(arctan(x) - arctan(y)) = tan(pi + arctan[(x-y)/(1+xy)])
(tan(arctan(x)) - tan(arctan(y)))/(1 + tan(arctan(x)) * tan(arctan(y))) = (tan(pi) + tan(arctan[(x-y)/(1+xy)]))/(1 + tan(pi) * tan(arctan[(x-y)/(1+xy)]))
(x - y)/(1 + xy) = (0 + [(x - y)/(1 + xy)])\(1 + (0)[(x + y)/(1 + xy)])
(x - y)/(1 + xy) = (x - y)/(1 + xy)