Question

need help with this one, need equation of tangent line at point (4,2) an the curve y +square root of x over y = 3 I don't even know who to go about getting this answe

Answer 1

for finding the equation of a line, you should first determine the slope of the line and one point of that line. so try to find the slope of the line and one point from that line

Answer 2

I am stuck finding the implicit derivative of\[y+(\sqrt{x}/y) =3\]

Answer 3

differentiate the equation and you will reach to the following relation:
dy+(0.5*(x^-0.5)*ydx-sqrt(x)*dy)/(y^2)=0, then replace the values of x and y in the equation and you wil find dy/dx.

Answer 4

ok i'm going to assume your trouble lies with:
\[(\frac{\sqrt{x}}{y})'=\frac{(\sqrt{x})'y-\sqrt{x} (y)'}{y^2}=\frac{\frac{1}{2 \sqrt{x}} y-\sqrt{x}y'}{y^2}\]
\[=\frac{2 \sqrt{x}}{2 \sqrt{x}} \cdot \frac{\frac{1}{2 \sqrt{x}} y-\sqrt{x}y'}{y^2}=\frac{y-2xy'}{2 \sqrt{x} y^2}\]

Answer 5

i use the quotient rule
then i use the chain rule a few times
---
the rest was algebra
i got rid of the compound fraction by multiplying by some kind of 1

Answer 6

wonderful, between both of those answers I think I understand. I then need to find the equation of the tangent line at the point (4,2)

Answer 7

so after solving for y' then plug in 4 for xand 2 for y

Answer 8

the dy/dx confuses me to no end!!

Answer 9

it is the same as y'
you can use this notation if you prefer

Answer 10

thank you all very much, that makes my day