## anonymous one year ago Derivative of inverse question.

1. anonymous

2. anonymous

What I did was I found the inverse by interchanging x and y and plugged in y=1 to solve for x. Is that correct?

3. anonymous

Ohh whoops forgot one step. I interchanged x and y, implicitely differentiated and then plugged in 1 for y.

4. zepdrix

Mmmm I'll share my thought process. Maybe not the way you're thinking of doing it, but it's just a thought. If you recall, the composition of a function and it's inverse gives back the argument,$\large\rm f(f^{-1}(x))=x$If we differentiate this equation, we get a nice formula for the derivative of an inverse function.

5. anonymous

Yeah I know that method. I was just trying to avoid using it because I don't like it personally :P .

6. zepdrix

By the chain rule,$\large\rm f'(f^{-1}(x))\cdot (f^{-1})'(x)=1$No bueno? :o Hmm ok i'll try to make sense of your madness you got there.

7. anonymous

8. zepdrix

my goodness.... that handwriting... does that say t plus costco?

9. Jhannybean

$f(x)=x^5+\sin(x) + 2x+1$$x=y^5+\sin(y)+2y+1$$\frac{d}{dx} \left(x=y^5+\sin(y)+2y+1\right)$$1=5y^4y' +y'\cos(y)+2y'+0$$1=y'(5y^4+\cos(y)+2)$$y'=\frac{1}{5y^4+\cos(y)+2}$

10. anonymous

$\frac{ 1 }{ 7+\cos(1) }$ better? :P .

11. Jhannybean

t + costco xD

12. anonymous

Cool! So if I plug in 1 I get the same answer with my methos :P .

13. anonymous

method*

14. Jhannybean

$y'(1) = \frac{1}{5(1)^4+\cos(1)+2} = \frac{1}{7+\cos(1)}\qquad \checkmark$

15. anonymous

Yay!

16. anonymous

So I was right all along :3 .

17. Jhannybean