Question

can someone explain to me Implicit Differentiation:
if y^3 = 25x^2, determine dx/dt when x = 5 and dy/dt = 1

Answer 1

these are the steps but i don't understand them :/
3y^2*dy/dt = 50x*dx/dt
When x = 5, y^3 = 25*5^2
y^3 = 625
y = cube root 625 = 8.55
3(8.55)^3*1 = 50*5*dx/dt
1875 = 250dx/dt
7.5 = dx/dt

Answer 2

isn't y=2.92402?

Answer 3

i did this on the calculator: sqrt (625)^(1/2)

Answer 4

(1/3)**

Answer 5

@satellite73

Answer 6

Let's start with this part...
\[\Large \frac{ d }{ dt } y^3\]use the chain rule to differentiate it... so bring down the exponent, reduce the exponent by one, then multiply by the derivative of y w. respect to t \[\Large 3 y^2 *\frac{ dy }{ dt }\]

Answer 7

so I can't replace the dy/dt by 1 since it's given?

Answer 8

@agent0smith

Answer 9

Well yes but i was seeing if you understood the differentiation... don't worry about plugging in numbers till later.

Answer 10

and 3y^2 is already the derivative, why do we need to use the chain rule?

Answer 11

@agent0smith

Answer 12

Because you have to multiply by the derivative of y with respect to t.

Answer 13

3y^2 is NOT the derivative of y^3, UNLESS you're just differentiating with respect to y.

Answer 14

normally we have y dependent on x
here y, and x both are dependent on some t, so we dont know what the function is
so instead of y^3 >>> 3y^2 we have the chain rule y^3 >>> 3y^2 * y'

Answer 15

It's the same with dy/dx... the derivative of y with respect to x is not just 1... it's 1*dy/dx

Answer 16

ok... :/ it's a little confusing..

Answer 17

agent0smith explained it nicely:
\[\frac{ d }{ dy }(y ^{3}) =3y ^{2}\]
But
\[\frac{ d }{ dt }y ^{3} = \frac{ d }{ dy }y ^{3}\frac{ dy }{ dt } = 3y ^{2}\frac{ dy }{ dt }\]

Answer 18

It's the same thing with the chain rule \[\huge \frac{ d }{ dx } f(x)^n = n*f(x)^{n-1} * f \prime (x)\]
you have to always multiply by the derivative at the end.

Answer 19

why can't we plug in the numbers right away if it's given?

Answer 20

You can... you're probably just better off not doing it until you really know what you're doing.

Answer 21

right i understand that... @ranga

Answer 22

well i was taught to plug in.. that's why.

Answer 23

that's the confusing part.

Answer 24

Fair enough... but you can't plug in until you have this step: 3y^2*dy/dt = 50x*dx/dt

Answer 25

exactly what i have right now.

Answer 26

so dy/dt is 1...

Answer 27

in the steps above, why is this: 3(8.55)^3*1 = 50*5*dx/dt?

Answer 28

y=8.55? y is suppose to be 2.92402.

Answer 29

and it's raised to the 3rd, and why not 2?

Answer 30

When x = 5, y^3 = 25*5^2
y^3 = 625
y = cube root 625 = 8.55

i think the next line after that is a mistake and it should by to the power of 2

Answer 31

so y=2.92402 is correct?

Answer 32

y^3 = 625
y = cube root 625 = 8.55

Answer 33

how?

Answer 34

i did this in the calculator: sort(625)^(1/3)

Answer 35

sqrt*

Answer 36

And you'll get 8.55... 2.94 cubed is close to 3^3 which is 27. Not 625.

Answer 37

i swear im getting 2.92402

Answer 38

i plugged it in exactly like iwrote it.

Answer 39

sort(625)^(1/3)

why are you taking the square root of a cubed root...

Answer 40

y^3 = 625
y = 625^(1/3)

Answer 41

OHHH

Answer 42

i thought the sqrt could do cube root.

Answer 43

if we did the exponent also.

Answer 44

That would mean ((625^(1/3))^(1/2) = 625^(1/6)

Answer 45

thank you! i got it !