Question

Hello I have a question on how to do this chain rule question

Answer 1

find \[\frac{ dy }{ dx }\] at x=0

\[y=\frac{ 1 }{ 1+u^2 }, u=2x+1\]

Answer 2

\[y=\frac{1}{1+(2x+1)^2}\]

yes?

Answer 3

Oh so you just plug in u for the equation. So then do I solve the derivitive of Y?

Answer 4

if so, rewrite using exponents...

\[y=\frac{1}{1+(2x+1)^2} = \left(1+(2x+1)^2\right)^{-1}\]

Answer 5

which you can use u... but you'll get this...

\[y=\left(1+u^2\right)^{-1}\Rightarrow \frac{d y}{d x}= -1\left(1+u^2\right)^{-2}\cdot \frac{d (1+u^2)}{dx}=-1\left(1+u^2\right)^{-2}\cdot2u\frac{du}{dx}\]

Answer 6

Ok so then I just solve the derivative like normal but with the chain rule

\[-1(1+(2x+1)^2)^{-2}(2(2x+1)(x+1))\]

Answer 7

and \[\frac{du}{dx}=\frac{d(2x+1)}{dx}=2\]

Answer 8

not quite...
\[\frac{dy}{dx}=\left(1+(2x+1)^2)\right)^{-2}\cdot2(2x+1)\cdot2=\frac{4(2x+1)}{\left(1+(2x+1)^2)\right)^{2}}\]

Answer 9

oh

Answer 10

you have to use the chain rule a couple of times. that will happen with functions that are nested like that.

Answer 11

oops, i had an extraneous ) in the denominator... please ignore it.

Answer 12

OK so I know the formula \[\frac{ dy }{ dx }=(dy/du)(du/dx)\]
So what does dy/du equal and du/dx equal