Question

Find dy/dx

Answer 1

\[\sqrt{x} = 5\sqrt{y}\]

Answer 2

why wouldn't it just be
\[\frac{\sqrt{x}}{5}=\sqrt{y}\to \frac{x}{25}=y\] and then
the derivative of y=ax?

Answer 3

well i'm supposed to multiply y by dy/dx

Answer 4

i got \[\frac{ 1 }{ 2\sqrt{x} }= 5 * \frac{ 1 }{ 2\sqrt{y} } *dy/dx\]

Answer 5

but idk... if that's right or if i'm going to the right direction

Answer 6

Ah, Implicitly. Ok
\[ (x)^{1/2}=5(y)^{1/2}\]
\[\frac{1}{2}x^{-1/2}=\frac{1}{2}5(y)^{-1/2}y^{\prime}\]

Answer 7

yes that's right for implicit D!

Answer 8

Divide both sides by \[y^{-1/2}\] gives you a \[y^{1/2}\] on the RHS which you know in terms of x:)

Answer 9

sorry LHS

Answer 10

so did i do it right so far?

Answer 11

Yes!

Answer 12

\[\frac{1}{2}\frac{\sqrt{y}}{\sqrt{x}}\frac{2}{5}=\frac{dy}{dx}\]
\[\frac{\sqrt{y}}{\sqrt{x}}\frac{1}{5}\frac{5}{5}=\frac{dy}{dx}\]
\[\frac{5\sqrt{y}}{\sqrt{x}}\frac{1}{25}=\frac{dy}{dx}\]
\[\frac{\sqrt{x}}{\sqrt{x}}\frac{1}{25}=\frac{dy}{dx}\]
\[\frac{1}{25}=\frac{dy}{dx}\]

Answer 13

1/5 *\[\sqrt{y/x}\]

Answer 14

Yes, but you can resubstitute for y since your original equation was
\[\sqrt{x} = 5\sqrt{y}\]
you can rearrange
\[\frac{1}{5}\frac{\sqrt{y}}{\sqrt{x}}\] to
\[\frac{5\sqrt{y}}{25\sqrt{x}}\]
\[ 5\sqrt{y} \to \sqrt{x}\] so
\[\frac{\sqrt{x}}{25\sqrt{x}}=\frac{1}{25}\]

Answer 15

oh, well is that the oversimplified version of the answer?

Answer 16

its the simplified version. Which would you rather be told/use that the derivative is some ratio of variables or 1/25?

Answer 17

well my teacher prefers the ratio.

Answer 18

Then that's the answer:)

Answer 19

You could ask your teacher why you shouldn't resub and get 1/25... maybe get some extra points etc etc.

Answer 20

when i divide the derivative of 5 squareroot of 5, i get x/y instead of y/x

Answer 21

no, she wrote the answer down and it's the ratio version

Answer 22

i am really confused on getting the ratio answer :(

Answer 23

you have to watch the signs of our exponents.
\[\frac{d}{dx}x^{\frac{1}{2}}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2x^{\frac{1}{2}}}\]

Answer 24

i got that derivative for x, but i'm not so sure about my y

Answer 25

\[\frac{d}{dx} 5y^{\frac{1}{2}}\to5\frac{d}{dx}y^{\frac{1}{2}}\to 5\frac{1}{2}y^{\left(\frac{1}{2}-1\right)}\frac{dy}{dx} \]

Answer 26

5* \[-1/2\sqrt{y}\]

Answer 27

why -(1/2) ? did you move the sqrt(y) to the LHS?

Answer 28

no because 1/2-1 is -1/2?

Answer 29

\[\frac{d}{dx}a^r = ra^{r-1}\]

Answer 30

yea... it's -1/2... if you don't leave negative exponents