Question

Find dy/dx for y=Sqrt(x(x+7))

Using logarithmic differentiation. Answer can only be in terms of x.

Thanks in advance for all the help yall have been giving me tonight! Learning much more than I do by going to class, lol!

Answer 1

Where I get particularly confused is how do I eliminate the 1/y variable from the left side of the equation that comes from taking the derivative of (lny)' if that helps anyone understand my problem!

Answer 2

No wait you can it's just tedious

Answer 3

Just plug in (x^2 + 7x)^(1/2) for y after you derive

Answer 4

And then put that back on the right side of the equation

Answer 5

\[y=\sqrt{x(x+7)} \\ \ln(y)=\ln([x(x+7)]^\frac{1}{2}) \\ \ln(y)=\frac{1}{2}(\ln(x(x+7)))\]
It would be totally cute to just write that ln(x(x+7)) as ln(x)+ln(x+7)

Answer 6

can you tell us how far you have gotten

Answer 7

like have you gotten to the differentiating part
if so can we see it

Answer 8

I have y'/y=.5(2x+7/(x^2+7x)) I'm pretty sure that's correct? Did it a little different than you though I distributed the x(x+7) before I did anything

Answer 9

looks fabulous

Answer 10

now do what @master50777 said

Answer 11

multiply the y on both sides

Answer 12

and then replace y with sqrt(x^2+7x)

Answer 13

OHH!!!!!!! It wasn't clicking with me where you guys were getting what y=... I'm silly hahaha. Thanks again @master50777 , and also to you @freckles !

Answer 14

yah a lot of people forget the initial equation and what it was called
not just you

Answer 15

Awesome to learn to look out for that now though!

Answer 16

you will see it some more in calculus
if you find y'' given an implicit equation for y

Answer 17

\[\text{ example } x^2+y^2=4 \\ \text{ find } y'' \\ 2x+2yy'=0 \\ y'=\frac{-2x}{2y}=\frac{-x}{y} \\ y''=\frac{(-1)y-y'(-x)}{y^2} \\=\frac{-y+y'x}{y^2} \text{ first replace } y' \text{ with } \frac{-x}{y} \\ y''=\frac{-y-\frac{-x}{y}(-x)}{y^2} =\frac{-y-\frac{x^2}{y}}{y^2} \text{ multiply both \top and bottom by } y \\ = \frac{-y^2-x^2}{y^3} =-\frac{x^2+y^2}{y^3} \\ \text{ but see initial equation } x^2+y^2=4 \text{ so we have } \\ y''=\frac{-4}{y^3}\]
just for fun
you can look at this or not

Answer 18

No I'll definitely look over that! My teacher doesn't quite explain in her lectures the way I need and I thank you guys a ton this is saving my retrice the morn I learn the better!

Answer 19

i bet you will assume be covering that topic if you haven't already

Answer 20

soon not assume
I'm sorry my first language is english but I don't like it sometimes :p

Answer 21

more** not morn. Your English is perfect pal