# Inverse Trig Derivatives

1. Jun 13, 2005

### Struggling

hi,
my problem is i have equations like determine the derivative of:

a. arcsin 1/x
b. arcsin x^(1/2)
c. arcsin x^(-1/2)
etc etc.

and i have the formulas for the trigs such as:

d/dx arcsinf(x) = 1/(1-f(x)^2)^1/2 x f'(x) and i have the arcos and asrctan one.

but when i put f(x) into the equations eg:

arcsin^(1/2):

1/ (1-x^1/2)^1/2 x 1/2x^-1/2

i can never get the answer i know im multiplying wrong and possibly using the powers wrong because and example the teacher used was:

y = arctan 3/x^2 find y'

f(x) = 3x^-2
f'(x) = -6x^-3

y'= 1/1+(3x^-2)^2 x (-6x^-3)

therefore: y' = -6x/ x^4 + 9

how did he get the final answer from y'= 1/1+(3x^-2)^2 x (-6x^-3)??? im totaly confused ive been trying to work it out for days any help would be appreciated thanks

2. Jun 13, 2005

### whozum

$$y = arctan\left(\frac{3}{x^2}\right)$$

$$\frac{d}{dx} arctan(x) = \frac{1}{1+x^2}}$$

By chain rule

$$\frac{d}{dx} arctan(f(x)) = \frac{f'(x)}{1+f(x)^2}}$$

$$f(x) = \frac{3}{x^2} \Rightarrow \frac{d}{dx} f(x) = \frac{-6}{x^3}$$

$$\frac{d}{dx} arctan\left(\frac{3}{x^2}\right) = \frac{\frac{-6}{x^3}}{1+\left(\frac{9}{x^4}\right)}$$ This is a correct answer, but can be simplified to:

$$\frac{-6}{x^3+\frac{9}{x}} = \frac{-6}{\frac{x^4+9}{x}} = \frac{-6x}{x^4+9}$$

3. Jun 13, 2005

### Struggling

i dont understand why it become x^4 here? i understand the whole thing up to this point?

4. Jun 13, 2005

### TenaliRaman

a+(b/c)
= (ac/c) + (b/c) .... (since c/c = 1)
= (ac+b)/c

-- AI

5. Jun 13, 2005

### whozum

$$\frac{d}{dx} arctan\left(\frac{3}{x^2}\right) = \frac{\frac{-6}{x^3}}{1+\left(\frac{9}{x^4}\right)}$$

$$\frac{\frac{-6}{x^3}}{1+\left(\frac{9}{x^4}\right)} \times \frac{x^3}{x^3} = \frac{\frac{-6}{x^3}x^3}{(1+\left(\frac{9}{x^4}\right))x^3} = \frac{-6}{x^3+\frac{9}{x}}$$

Then make a common denominator for the bottom, [itex] x^3 = x^4/x [/tex]

$$\frac{-6}{x^3+\frac{9}{x}} = \frac{-6}{\frac{x^4}{x} + \frac{9}{x}}.$$

Combine fractiosn with a common denominator.

$$\frac{-6}{\frac{x^4+9}{x}}$$

Dividing is the same as multiplying by the reciprocal.

$$\frac{-6}{\frac{x^4+9}{x}} = -6 \times \frac{x}{x^4+9}$$ and you get the final result. This is basic algebra.