# Thread: inverse function

1. ## inverse function

f(x)=tan^-1 (-x/x+1)

show that f'(x)=-1/2x^2+2x+1

using the formula tan^-1 x/a =a/a^2+x^2

f'(x)=-x/(x+1)^2 +(-x)^2

sadly this is not right

2. ## Re: inverse function

Originally Posted by markosheehan
f(x)=tan^-1 (-x/x+1)
-x/x+ 1= 0 for all x. I am sure you mean tan^1(-x/(x+ 1)).

show that f'(x)=-1/2x^2+2x+1
and here you mean f'(x)= -1/(2x^2+ 2x+ 1), right?

using the formula tan^-1 x/a =a/a^2+x^2
You mean the derivative of the arctan of x/a is equal to that, not the arctan itself.
The derivative of tan^1 x= 1/(1+ x^2) (Please use parentheses!) so first replace that "x" by "-x/(x+ 1)":
1/(1+ x^2/(x+ 1)^2).
Simplify that by multiplying numerator and denominator by (x+ 1)^2: (x+1)^1/((x+1)^2+ x^2= (x+ 1)^2/(2x^2+ 2x+ 1).

Now, using the chain rule, multiply that by the derivative of -x/(x+ 1). Using the quotient rule, that is [-(x+ 1)- 1(-x)]/(x+ 1)^2= -1/(x+ 1)^2 to get
[(x+ 1)^2/(2x^2+ 2x+ 1)][-1/(x+ 1)^2]= -1/(2x^2+ 2x+ 1).

f'(x)=-x/(x+1)^2 +(-x)^2

sadly this is not right

3. ## Re: inverse function

Originally Posted by HallsofIvy
-x/x+ 1= 0 for all x. I am sure you mean tan^1(-x/(x+ 1)). . . . . . tan^-1(-x/(x+ 1))

The derivative of tan^1 x= 1/(1+ x^2) . . . . . tan^-1(x) = 1/(1+ x^2)

Simplify that by multiplying numerator and denominator by (x+ 1)^2:

(x+1)^1/((x+1)^2+ x^2 . . . . . (x+1)^1/((x+1)^2+ x^2)

= (x+ 1)^2/(2x^2+ 2x+ 1).
.

4. ## Re: inverse function

yes you have read my post right.

i understand you up to the point where you say
"Now, using the chain rule, multiply that by the derivative of -x/(x+ 1). Using the quotient rule, that is [-(x+ 1)- 1(-x)]/(x+ 1)^2= -1/(x+ 1)^2 to get
[(x+ 1)^2/(2x^2+ 2x+ 1)][-1/(x+ 1)^2]= -1/(2x^2+ 2x+ 1)."

i do not see why we have to use the chain rule. the formula tan^1 x= 1/(1+ x^2) is supposed to give you the derivative it for you i thought. i do not see why you multiply by the derivative of -x/(x+ 1)

5. ## Re: inverse function

$y=\arctan(\frac{-x}{x+1})$

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

Let $u(x)= \frac{-x}{x+1}$, with $\frac{du}{dx}=\frac{-(x+1)+x}{(x+1)^2}=\frac{-1}{(x+1)^2}$ (Quotient rule).

Now $y(u)=\arctan(u)$ and $\frac{dy}{du}=\frac{1}{1+u^2}$ which is given in the question.

By the chain rule, $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

Can you finish it from there?