Question

help please!

Answer 1

if w=\[e^{x ^{2}} and x=\frac{ t+1 }{ t-1 }, find \frac{ dw }{ dt } when x=2\]

Answer 2

Would you understand how to find dw/dx using chain rule?\[\large\rm w=e^{x^2}\]\[\large\rm w'=?\]

Answer 3

yes

Answer 4

:( sorry saw the msg late. hope to catch you again

Answer 5

Then what is it? :U

Answer 6

\[\large\rm \frac{dw}{dx}=e^{x^2}(2x)\]Ok great.
If we take the derivative of w, with respect to some other variable,
then our derivative will look very very similar,\[\large\rm \frac{dw}{dt}=e^{x^2}(2x)x'\]

Answer 7

So as some side work,
we'll have to figure out what the derivative of x, with respect to t, is.

Answer 8

\[\large\rm x=\frac{t+1}{t-1},\qquad\qquad\qquad x'=?\]Quotient rule, ya?

Answer 9

okay so i use the quotient rule for that?

Answer 10

yes let me find that

Answer 11

\[\frac{ -2 }{ (t-2)^{2} }\]

Answer 12

\[\large\rm \frac{dw}{dt}=e^{x^2}(2x)x'\]Ok great,\[\large\rm \frac{dw}{dt}=e^{x^2}(2x)\left[\frac{-2}{(t-1)^2}\right]\]

Answer 13

Ohh, we have some more side work to do :(
When x=2, we need to determine what t=.\[\large\rm 2=\frac{t+1}{t-1}\]Solve for t.

Answer 14

ok

Answer 15

3

Answer 16

Oooo k great!
So plug in x=2, t=3 and bammmm we got it!

Answer 17

plug where?

Answer 18

This is our derivative that we've established,\[\large\rm \frac{dw}{dt}=e^{x^2}(2x)\left[\frac{-2}{(t-1)^2}\right]\]They want you to evaluate the derivative when x=2 and t=3.

Answer 19

wwops i made a mistake

Answer 20

they when t=2 not x=2

Answer 21

Oh :o

Answer 22

Then figure out x when t=2,\[\large\rm x=\frac{t+1}{t-1}\]

Answer 23

im not sure what they want u to do here

Answer 24

x=3