Mathematics OpenStudy (anonymous):

i need help finding the linear approximation of f(x,y)=x^2/(y^2+1) OpenStudy (anonymous):

local linear approximation? OpenStudy (anonymous):

Linear approximation usually involve a given point. Please write in question word for word as given. OpenStudy (anonymous):

let f be defined as follows: f(x,y)=x^2/(y^2+1). use the linear approximation at an appropriate point (a,b) to estimate f(6.03,0.95). OpenStudy (anonymous):

you there? OpenStudy (anonymous): OpenStudy (anonymous):

are you able to help me solve this? OpenStudy (anonymous):

just hold tight a sec. OpenStudy (anonymous):

k OpenStudy (anonymous):

\[L(x,y)=f (x _{0},y _{0})+f _{x}(x _{0},y _{0})(x -x _{0})+f _{y}(x _{0},y _{0})(y -y _{0})\] OpenStudy (anonymous):

and once i plug in those values.. that would be my answer? OpenStudy (anonymous):

Yes. be careful, easy to make mistakes. Note partial derivative in respect to x, partial to y. In the case\[(x-x _{0})\]and y\[(y-y _{0})\]you keep the actual letter x and the letter y, where ever you see\[x _{0},y _{0}\]you plug in given x and y. OpenStudy (anonymous):

ok makes sense... but the answer is asking for a decimal answer... it wont take the whole thing. OpenStudy (anonymous):

any suggestions for this? OpenStudy (anonymous):

After re-reading question, they say an appropriate point (a,b). They want you to read between the lines and know the appropriate (a,b) is (6.00, 1.00). So above, where I say keep actual letter x and y, instead plug in x=6.00, y=1.00 OpenStudy (anonymous):

o ok i see what they are saying.. i will try this again and keep my fingers crossed. thank you.

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