If 2x2 + y2 = 17 then evaluate the second derivative of y with respect to x when x = 2 and y = 3. Round your answer to 2 decimal places. Use the hyphen symbol, -, for negative values.

Answer:y''=-1.26Step-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 xThen , 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=3Then, 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