Differentiation of Inverse Trigonometric Functions

[sharp]

Hey guys, I have to know how to Differentiate Inverse Trigonometric Functions in my next exam and need somewhere to study up on them. Do you know of any web sites I could read? Can't find anything on Karl's Calculus.

Thanks

 

[finchie_88]

Well, the integral of $$\frac{1}{ a^2 + x^2 }$$ is $$\frac{1}{a}tan^{-1}(x/a)$$. There are a few more standard ones for inverse sin and cos, but can't remember them off the top of my head.

Edit: I think its $$\frac{1}{\sqrt{a^2 - x^2}}$$ for $$sin^{-1}(x/a), and the negative of that for inverse cos, but don't quote me on that.$$

 

[nrqed]

Hey guys, I have to know how to Differentiate Inverse Trigonometric Functions in my next exam and need somewhere to study up on them. Do you know of any web sites I could read? Can't find anything on Karl's Calculus.

Thanks

Consider for example y =arcsin(x).

The way to do this is to rewrite this as sin(y) =x and to differentiate both sides so that cos(y) dy = dx, so you get dy/dx= 1/cos(y)

Now, you need to rewrite cos(y) in terms of x. The trick is this: we started from y =arcsin(x) or, equivalently, sin(y) =x. So think of y as an angle in arigt angle triangle. Then x would have to be the side opposite to that angle divided by the hypothenuse of the triangle, right?

so y = angle and x = opposite side/hypothenuse.

Just with that we cannot fix the opposite side or the hypothenuse separately but might as well take the easiest possibility: let's say that the opposite side *is* x and that the hypothenuse is equal to one.
So we have a right angle triangle with an hypothenuse of 1 and one side of length x and an angle y which is opposite to the side of length x. So far so good?

Now, what is cos (y) in that triangle? It is the adjacent side to the angle y divided by the hypothenuse. The hypothenuse is 1 and the adjacent side is obviously sqrt(hyp^2 -opp^2) = sqrt(1 - x^2).
So cos(y) =sqrt(1-x^2)

so finally, dy/dx = 1/cos(y) = 1/sqrt(1-x^2)

You can look up the results for any other inverse trig function and rederive them using the same approach.

Hope this helps.

Patrick

 

[Curious3141]

Consider for example y =arcsin(x).

The way to do this is to rewrite this as sin(y) =x and to differentiate both sides so that cos(y) dy = dx, so you get dy/dx= 1/cos(y)

Now, you need to rewrite cos(y) in terms of x. The trick is this: we started from y =arcsin(x) or, equivalently, sin(y) =x. So think of y as an angle in arigt angle triangle. Then x would have to be the side opposite to that angle divided by the hypothenuse of the triangle, right?

so y = angle and x = opposite side/hypothenuse.

Just with that we cannot fix the opposite side or the hypothenuse separately but might as well take the easiest possibility: let's say that the opposite side *is* x and that the hypothenuse is equal to one.
So we have a right angle triangle with an hypothenuse of 1 and one side of length x and an angle y which is opposite to the side of length x. So far so good?

Now, what is cos (y) in that triangle? It is the adjacent side to the angle y divided by the hypothenuse. The hypothenuse is 1 and the adjacent side is obviously sqrt(hyp^2 -opp^2) = sqrt(1 - x^2).
So cos(y) =sqrt(1-x^2)

so finally, dy/dx = 1/cos(y) = 1/sqrt(1-x^2)

You can look up the results for any other inverse trig function and rederive them using the same approach.

Hope this helps.

Patrick

Just to add on to Patrick's excellent post, there are six basic and derived circular inverse trig ratios (arcsin, arccos, arctan, arccosec, arcsec, arccot). But you don't have to remember all of them. Just remember arcsin, arccosec and arctan and the rest can be easily derived from them.

Observe that (using the first quadrant domain),

$$\arcsin(x) + \arccos(x) = arccsc(x) + arcsec(x) = \arctan(x) + arccot(x) = <br /> \frac{\pi}{2}$$

So when you have differentiated y = arcsin(x) to get y',

$$z = \arccos(x) = \frac{\pi}{2} - \arcsin(x)$$

$$z' = -y'$$

and so forth for the other inverse trig ratios.

 

[arildno]

I've never used the derivative of arccosecant.

 

[Curious3141]

I've never used the derivative of arccosecant.

Well, evaluate $$\int{-\frac{1}{x\sqrt{x^2-1}}}dx$$, then.