anonymous
  • anonymous
Derivative of inverse question.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • 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?
anonymous
  • anonymous
Ohh whoops forgot one step. I interchanged x and y, implicitely differentiated and then plugged in 1 for y.

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zepdrix
  • 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.
anonymous
  • anonymous
Yeah I know that method. I was just trying to avoid using it because I don't like it personally :P .
zepdrix
  • 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.
anonymous
  • anonymous
|dw:1444463152370:dw| is my final answer.
zepdrix
  • zepdrix
my goodness.... that handwriting... does that say `t plus costco`?
Jhannybean
  • Jhannybean
\[f(x)=x^5+\sin(x) + 2x+1 \]\[x=y^5+\sin(y)+2y+1\]\[\frac{d}{dx} \left(x=y^5+\sin(y)+2y+1\right)\]\[1=5y^4y' +y'\cos(y)+2y'+0\]\[1=y'(5y^4+\cos(y)+2)\]\[y'=\frac{1}{5y^4+\cos(y)+2}\]
anonymous
  • anonymous
\[\frac{ 1 }{ 7+\cos(1) }\] better? :P .
Jhannybean
  • Jhannybean
t + costco xD
anonymous
  • anonymous
Cool! So if I plug in 1 I get the same answer with my methos :P .
anonymous
  • anonymous
method*
Jhannybean
  • Jhannybean
\[y'(1) = \frac{1}{5(1)^4+\cos(1)+2} = \frac{1}{7+\cos(1)}\qquad \checkmark \]
anonymous
  • anonymous
Yay!
anonymous
  • anonymous
So I was right all along :3 .
Jhannybean
  • Jhannybean
Yep, trust your gut feeling.

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