Question

Derivative of inverse question.

Answer 1

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?

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

is my final answer.

Answer 7

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

Answer 8

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

Answer 9

\[\frac{ 1 }{ 7+\cos(1) }\]

better? :P .

Answer 10

t + costco xD

Answer 11

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

Answer 12

method*