Question

will award medal
Find the linear approximation of the given function at (2π, 0).
f(x,y)=sqrt(y + (cos(x))^2
@SolomonZelman

Answer 1

don't get the function

Answer 2

\[f(x,y)=\sqrt{y+(\cos(x))^2}\]

Answer 3

then you left out a bracket. but good that you fixed it.

Answer 4

so my partial with respect to x i think is \[\frac{ 2\cos(x)(-\sin(x)) }{ 2\sqrt{y+(\cos(x))^2} }\]

Answer 5

looks fine, the y on top those is 0?

Answer 6

derivative of y is 1, no ?

Answer 7

the chain for cos^2(x) is good

Answer 8

no because the derivitive of y with respect to x is zero, because y is a constant

Answer 9

or treated as one

Answer 10

oh, ttrruuee

Answer 11

yes, it is just not something I used to though:)

Answer 12

Yes.

Answer 13

but for my derivitive with respect to y, would you use the chain rule for (cos(x))^2, and instead of f(x)=x^2 and g(x)=cos(x)... f(y)=y^2, g(y)=cos(x)?

Answer 14

if you are differentiating with respect to y, then the cos(x) is constant.

Answer 15

and so is cos^2x

Answer 16

so f_y would just be 1?

Answer 17

no

Answer 18

the square root has a variable inside (the y), so differentiate the root, and apply chain

Answer 19

it would be sqrtx

Answer 20

I have to go right now to hear a guest speaker in my community. Excuse me

Answer 21

im not stalking anyone... you have no proof

Answer 22

1/(sqrt of this) times derivative of what is in the root.

Answer 23

I mean 1/(2( sqrt of that ))

Answer 24

i gogt to go... I will be back in an hour.... sorry again.

Answer 25

\[f(x,y)=\sqrt{y+\cos^2x}~~\implies~~\nabla f(x,y)=\left(-\frac{\cos x\sin x}{\sqrt{y+\cos^2x}},\,\frac{1}{2\sqrt{y+\cos^2x}}\right)\]
At \((2\pi,0)\), you have \(\nabla f=\left(0,\dfrac{1}{2}\right)\), and so the linear approximation is given by
\[z-f(2\pi,0)=0(x-2\pi)+\frac{1}{2}(y-0)\]

Answer 26

Let me know if that code is showing up again.

Answer 27

yea it was

Answer 28

could you do the partial with respect to y step by step?

Answer 29

@SithsAndGiggles

Answer 30

\[f(x,y)=\sqrt{y+\cos^2x}\]
\[f_y=\frac{1}{2}(y+\cos^2x)^{-1/2}\frac{\partial }{\partial y}[y+\cos^2x]=\frac{1}{2\sqrt{y+\cos^2x}}\]

Answer 31

That is, power rule and chain rule.