Question

1) Find dy/dx ]x=4
if y=u^2-2u^5 and u=x-sqrt x.
2) Find dy/dx]t=1
if y=sqrt(1+r^2) and r= (t+1)/(2t+1)

Answer 1

this is chain rule
\[\frac{dy}{dx} = \frac{dy}{du}*\frac{du}{dx}\]

Answer 2

okay

Answer 3

do i sub in the x value into the u equation?

Answer 4

or do the chain rule first?

Answer 5

@freckles
Can you help me please?

Answer 6

@ganeshie8

Answer 7

` y=u^2-2u^5 and u=x-sqrt x.`

\[\frac{dy}{dx} = \frac{dy}{du}*\frac{du}{dx}\]

first find \(\dfrac{dy}{du}\) and \(\dfrac{du}{dx}\)

Answer 8

\(y = u^2 - 2u^5\)

\(\dfrac{dy}{du} = ?\)

Answer 9

(2u-10u^4)

Answer 10

\(u = x-\sqrt{x}\)

\(\dfrac{du}{dx} = ?\)

Answer 11

(1-1/2x^-1/2)

Answer 12

so by chain rule we have

\[\begin{align}\frac{dy}{dx} &= \frac{dy}{du}*\frac{du}{dx}\\~\\
&= \left(2u-10u^4\right)*\left(1-\frac{1}{2}x^{-1/2}\right)

\end{align}\]

Answer 13

you want to find the derivative at x = 4

so plugin x = 4 above

Answer 14

okay, so the answer would be -116?

Answer 15

try again

Answer 16

I'm getting -117

Answer 17

just double check once..

Answer 18

oh i forgot to put brackets, i got the -117 too.

Answer 19

me too! :)

Answer 20

wolfram too
http://www.wolframalpha.com/input/?i=%28%28x-sqrt%28x%29%29%5E2-2%28x-sqrt%28x%29%29%5E5%29%27+at+x%3D4

Answer 21

does 2 have type-o?

Answer 22

dy/dt]t=1 ?

Answer 23

yes, that' what i mean. thanks

Answer 24

So first we do the chain rule?

Answer 25

yah

Answer 26

\[\frac{dy}{dt}=\frac{dr}{dt} \cdot \frac{dy}{dr}\]

Answer 27

is this correct?
y' = 1/2 (1+r^2)(2r) ((-1(2t+1)(2))

Answer 28

\(y=\sqrt{1+r^2}\)

\(\dfrac{dy}{dr} = ?\)

Answer 29

oops... is it 1/2 (1+r^2)^-1/2 (2r)

Answer 30

Yes!

\( r= \dfrac{t+1}{2t+1}\)

\(\dfrac{dr}{dt} = ?\)


(you need to use quotient rule)

Answer 31

heard of quotient rule before ?
if not, you may use product rule also by writing 1/(2t+1) as (2t+1)^(-1)

Answer 32

I know the quotient rule, is it faster to use that?

Answer 33

both are same
but i would use qyotient rule here..

Answer 34

because i see a quotient here..

Answer 35

\[\large \left(\dfrac{u}{\color{blue}{v}}\right)' = \dfrac{\color{blue}{v}u' - u\color{blue}{v}'}{\color{blue}{v}^2}\]

Answer 36

is this correct?
\[\frac{ t+1-(t+1)(-2(2t+1)) }{ (2t+1)^2 }\]

Answer 37

Nope. numerator has a mistake
you need to start numerator with the term in denominator :

\[\large \left(\dfrac{u}{\color{blue}{v}}\right)' = \dfrac{\color{blue}{\bbox[4px,border:2px solid red]{v}}u' - u\color{blue}{v}'}{\color{blue}{\bbox[4px,border:2px solid red]{v}}^2}\]

\(\)

Answer 38

\[\begin{align} \left(\dfrac{t+1}{\color{blue}{2t+1}}\right)' &= \dfrac{\color{blue}{\bbox[4px,border:2px solid red]{(2t+1)}}(t+1)' - (t+1)\color{blue}{(2t+1)}'}{\color{blue}{\bbox[4px,border:2px solid red]{(2t+1)}}^2}\\~\\~\\~\\~\\

&= \dfrac{\color{blue}{(2t+1)}(1) - (t+1)\color{blue}{(2)}}{\color{blue}{(2t+1)}^2}\\~\\~\\~\\
&= \dfrac{\color{blue}{2t+1} - 2t-2}{\color{blue}{(2t+1)}^2}~\\~\\~\\~\\
&= \dfrac{-1}{\color{blue}{(2t+1)}^2}~\\~\\~\\~\\

\end{align}\]

Answer 39

so we have below :

\[\dfrac{dy}{dr} =\frac{1}{2} (1+r^2)^{-1/2} (2r) \]
\[\dfrac{dr}{dt} =\frac{-1}{(2t+1)^2} \]


multiply both and obtain \(\dfrac{dy}{dt}\)

Answer 40

\[\dfrac{dy}{dt} = \dfrac{dy}{dr}*\dfrac{dr}{dt}\]

Answer 41

\[\frac{ -2r }{ 2\sqrt{1+r^2}(2t+1)^2 }\]

Answer 42

@ganeshie8