Question

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 1

Answer:

y''=-1.26

Step-by-step explanation:

We are given that $2x^2+y^2=17$

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

$4x+2yy'=0$

$2x+yy'=0$

$yy'=-2x$

$y'=-\frac{2x}{y}$

Again differentiate w.r.t.x

Then , we get

$2+(y')^2+yy''=0$ $(u\cdot v)'=u'v+v'u)$

$2+(y')^2+yy''=0$

Using value of y'

$yy''=-2-(-\frac{2x}{y})^2$

$y''=-\frac{2+(-\frac{2x}{y})^2}{y}$

Substitute x=2 and y=3

Then, we get $y''=-\frac{2+(\frac{4}{3})^2}{3}$

$y''=-\frac{18+16}{9\times 3}=-\frac{34}{27}$

Hence,y''=-1.26