Calculus. Derivatives of inverse trig functions

Thread Starter


Joined Sep 20, 2009
Hello everyone, I'm having some trouble with the following problem. I'm not too sure how to attack it.

y = 10(cosx)^-1-10x(sech)^-1

Find Y`

I want to set this up as 10 d/dx (cosx)^-1 - 10d/dx xsech^-1
but I can't seem to come to the same answer my professor did -10 (sechx)^-1

Any tips?


Joined Feb 24, 2006
The way you are writing the function implies something I don't think you mean. When using what looks like an exponent to mean the inverse of a function it needs to go BETWEEN the function name and the argument:

\(10 cos^{-1}(x)\)

This reads as: "10 times the inverse cosine of x"

10x(sech)^-1 is not a meaningful expression.

Checkout this page for the derivation method for the inverse cosine.


Joined Jun 17, 2014
Hello there,

The interpretation of:
"I want to set this up as 10 d/dx (cosx)^-1 - 10d/dx xsech^-1"

although it was not written as clear as it should be is as follows due to the title that mentions 'inverse trig functions':

"I want to set this up as 10*d(acos(x))/dx - 10*d(x*asech(x))/dx"

In order to preform this calculation you can discard the 10 temporarily as that carries through due to the constant rule: d(K*f(x))/dx=K*d(f(x))/dx
You can then put it back later.
That leaves us with:
d(acos(x))/dx - d(x*asech(x))/dx

To solve this all you have to do is look up the inverse trig form derivatives and note that you also have to know how to do the x*f(x) form of a derivative. So it's just the two derivatives and perform the subtraction.
Then, simplify the result and compare to the known answer which without the 10 is just -asech(x). You can then put the 10's back if you want, or do that before you do the comparison.

So the main points are:
1. Look up the inverse trig form derivatives.
2. Note how to do the x*f(x) derivative (multiplication of functions of x).
3. Simplify and compare.
4. Learn how to properly write out algebraic and trig forms in text so there is a universal understanding, noting that cos(x)^-1 can be interpreted as either acos(x) or 1/cos(x) so nobody will know which one you mean unless you clarify it. You did specify in the title that you were dealing with inverse trig functions so that helped, but it's better to write it out in a more clear way as well.
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Thread Starter


Joined Sep 20, 2009
Sorry for the late reply everyone, long long day between work and class today. Thank you for your help. First off, as you've all noted I incorrectly typed out the equation, it is in fact arccos and arcsech. I'll take some time tomorrow to try the tips MrAl, I'll post the work at some point. Thanks again.