How can I simplify inverse trig derivatives with fractions?

In summary, the student is trying to solve for y' in terms of f(x) and x^4, but is getting stuck on x^4. They use the chain rule to simplify the equation to x^3.
  • #1
Struggling
52
0
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 I am 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)? I am totaly confused I've been trying to work it out for days any help would be appreciated thanks
 
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  • #2
[tex] y = arctan\left(\frac{3}{x^2}\right) [/tex]

[tex] \frac{d}{dx} arctan(x) = \frac{1}{1+x^2}} [/tex]

By chain rule

[tex] \frac{d}{dx} arctan(f(x)) = \frac{f'(x)}{1+f(x)^2}} [/tex]

[tex] f(x) = \frac{3}{x^2} \Rightarrow \frac{d}{dx} f(x) = \frac{-6}{x^3}[/tex]

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

[tex] \frac{-6}{x^3+\frac{9}{x}} = \frac{-6}{\frac{x^4+9}{x}} = \frac{-6x}{x^4+9}[/tex]
 
  • #3
whozum said:
[tex] \frac{-6}{x^3+\frac{9}{x}} = \frac{-6}{\frac{x^4+9}{x}} = \frac{-6x}{x^4+9}[/tex]

i don't understand why it become x^4 here? i understand the whole thing up to this point?
 
  • #4
a+(b/c)
= (ac/c) + (b/c) ... (since c/c = 1)
= (ac+b)/c

-- AI
 
  • #5
[tex] \frac{d}{dx} arctan\left(\frac{3}{x^2}\right) = \frac{\frac{-6}{x^3}}{1+\left(\frac{9}{x^4}\right)} [/tex]

[tex] \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}}[/tex]

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

[tex] \frac{-6}{x^3+\frac{9}{x}} = \frac{-6}{\frac{x^4}{x} + \frac{9}{x}}. [/tex]

Combine fractiosn with a common denominator.

[tex] \frac{-6}{\frac{x^4+9}{x}} [/tex]

Dividing is the same as multiplying by the reciprocal.

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

Question 1: What are inverse trig derivatives?

Inverse trig derivatives are mathematical functions that represent the rate of change of inverse trigonometric functions. They are used to find the slope of a curve at any given point on the inverse trigonometric function.

Question 2: How do you find the derivative of an inverse trig function?

To find the derivative of an inverse trig function, you can use the chain rule. First, you take the derivative of the inverse trig function itself, and then multiply it by the derivative of the inside function.

Question 3: What is the derivative of arctan x?

The derivative of arctan x is 1/(1+x^2). This can be derived using the quotient rule and the derivative of the inverse function.

Question 4: Are there any special rules for finding the derivatives of inverse trig functions?

Yes, there are special rules for finding the derivatives of inverse trig functions. These include the chain rule, the power rule, and the quotient rule. It is important to familiarize yourself with these rules in order to successfully find the derivative of an inverse trig function.

Question 5: Why are inverse trig derivatives important in science?

Inverse trig derivatives are important in science because they are used to solve a variety of problems in fields such as physics, engineering, and statistics. They help us understand the relationship between variables and their rates of change, which is crucial in many scientific applications.

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