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My question is, how can I solve the following derivative question?

$$y = \sin(\arctan x) + \tan(\arcsin x)$$

Thanks in advance

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@Isaac Thanks for your answer. – MAxcoder Dec 25 '10 at 22:36
You can comment on Isaac's answer instead of posting a new answer. – Jonas Meyer Dec 27 '10 at 2:13
up vote 9 down vote accepted

It's not clear to me where the derivative comes in to your question, but to simplify $\sin(\arctan x)+\tan(\arcsin x)$, I'd suggest the following: since arctangent and arcsine both result in angles in the first or fourth quadrants (between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$), draw two diagrams showing those two quadrants, one showing an angle labeled $\arctan x$, the other showing an angle labeled $\arcsin x$. Draw right triangles to the $x$-axis in each diagram and label the legs in terms of $x$; for example, if the angle is $\arctan x$, that suggests that the leg opposite the angle is $x$ and the leg adjacent to the angle is $1$ (make sure that the labeling makes sense for positive and negative values of $x$). Use these diagrams to find simpler algebraic expressions for $\sin(\arctan x)$ and $\tan(\arcsin x)$.

edit: Here's a bit more detail on part of the problem. If you're still stuck, please comment and let me know.

Start by drawing an angle in the proper quadrant(s). $\arctan x$ is an angle in the first or fourth quadrant, so:

diagram 1

($\tan^{-1}(x)$ is another notation for $\arctan x$.) Either the black (1st quadrant) or the gray (4th quadrant) triangle could be the proper triangle, depending on whether $x$ is positive or negative. Since the angle is $\arctan x$, the tangent of that angle is $x$. The easiest way to make that happen is for the leg opposite the angle to have length $x$ and the leg adjacent to the angle to have length $1$. Note that if $x$ is negative, the gray triangle is the correct triangle and in the sense of directed-distances in the coordinate plane, that leg should have "negative" length.

diagram 2

Knowing the lengths of 2 sides of a right triangle, you can use the Pythagorean theorem to find the length of the third side. In this case, the unknown side is the hypotenuse and the directed-distance length of the hypotenuse should always be positive.

diagram 3

With all three sides of the triangle labeled, you can use the triangle to write an expression for $\sin(\arctan x)$, which is the sine of the marked angle, so the ratio of the length of the opposite leg to the length of the hypotenuse, $\frac{x}{\sqrt{x^2+1}}$.

The procedure for the $\tan(\arcsin x)$ term is similar, but the labeling of the sides of the triangle is different.

If you were working with, for example, arccosine, the two triangles would be in the first and second quadrants because the range of the arccosine function is $0\le\arccos x\le\pi$.

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Approach with function simplifying must take into account the domain of the original function. If finding of the derivative is a part of another problem (for example, find the domain of differentiability), providing the simplified function derivative might give wrong answer. Most obvious example is y(x)=x/x. Simplifying it gives y(x)=1, which derivative is y'(x)=0, but the original function derivative is y'(x)=0 except x=0. – mbaitoff Jan 16 '11 at 7:27
@mbaitoff: in this case, the domain restriction(s) inherited from the original trigonometric and inverse trigonometric functions are carried over to the purely-algebraic expressions in an obvious way. – Isaac Jan 16 '11 at 8:07
@Isaac I didn't even consider trying this. Thank you for this explanation, this simplifies everything. – Korgan Rivera Aug 25 '12 at 4:43
how to did you make the pictures/ – yiyi Nov 18 '13 at 7:07
@yiyi: I don't remember for sure, but probably in Mathematica. – Isaac Nov 18 '13 at 17:32

First you want to simplify $\tan (\arcsin x)$ and $\sin(\arctan x)$. Let me show you the latter one.

Suppose $\theta=\arctan x$. Then by definition $-\pi/2<\theta<\pi/2$ and $\tan\theta=x$. Since $\sec \theta>0$ (because $-\pi/2<\theta<\pi/2$), and $$\sec^2\theta=1+\tan^2 \theta =1+x^2,$$ you get $\sec \theta=\sqrt{1+x^2}$, so $\cos \theta =1/\sqrt{1+x^2},$ and $$\sin\theta=\tan\theta \cos\theta =\frac{x}{\sqrt{1+x^2}},$$ i.e. $\sin(\arctan x)=\frac{x}{\sqrt{1+x^2}}$.

Similarly, you can prove $\tan (\arcsin x)=\frac{x}{\sqrt{1-x^2}}.$

Using quotient rule, you can prove that the derivative of $\frac{x}{\sqrt{1+x^2}}$ and $\frac{x}{\sqrt{1-x^2}}$ are

$$\frac{1}{(1+x^2)^{3/2}} \mbox{ and }\frac{1}{(1-x^2)^{3/2}} $$

respectively. Hence the derivative of your function is $$\frac{1}{(1+x^2)^{3/2}}+\frac{1}{(1-x^2)^{3/2}}. $$

${\bf Edit.}$ Or you can use chain rule to get:

$$\cos(\arctan x)\frac{1}{1+x^2}+\sec^2(\arcsin x)\frac{1}{\sqrt{1-x^2}}$$

Now simplify it.

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What's your intended result?

A derivative, i.e. $y'$? Use the chain rule.

Solve for $x$? Use $\cos^2(x)=1-\sin^2(x)$ and $\tan(x)=\sin(x)/\cos(x)$ to simplify

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