Answer:
y''=-1.26
Step-by-step explanation:
We are given that [tex]2x^2+y^2=17[/tex]
We have evaluate the second order derivative of y w.r.t. x when x=2 and y=3.
Differentiate w.r.t x
Then , we get
[tex]4x+2yy'=0[/tex]
[tex]2x+yy'=0[/tex]
[tex]yy'=-2x[/tex]
[tex]y'=-\frac{2x}{y}[/tex]
Again differentiate w.r.t.x
Then , we get
[tex]2+(y')^2+yy''=0[/tex] [tex](u\cdot v)'=u'v+v'u)[/tex]
[tex]2+(y')^2+yy''=0[/tex]
Using value of y'
[tex]yy''=-2-(-\frac{2x}{y})^2[/tex]
[tex]y''=-\frac{2+(-\frac{2x}{y})^2}{y}[/tex]
Substitute x=2 and y=3
Then, we get [tex]y''=-\frac{2+(\frac{4}{3})^2}{3}[/tex]
[tex]y''=-\frac{18+16}{9\times 3}=-\frac{34}{27}[/tex]
Hence,y''=-1.26
