Question

If 3x2 + y2 = 7 then evaluate the second derivative of y with respect to x when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -, for negative values.

Answer 1

Do you think you can help? @surjithayer

Answer 2

\[3x^2+y^2=7\]
diff. w.r.t . x
\[6x+2y \frac{ dy }{ dx }=0,\frac{ dy }{ dx }=\frac{ -3x }{ y }\]
\[\frac{ d^2y }{ dx^2 }=\frac{ y \left( -3 \right)-\left( -3x \right)\frac{ dy }{ dx } }{ y^2 }=\frac{ -3y+3x \left( \frac{ -3x }{ y } \right) }{ y^2 }\]
\[\frac{ d^2y }{ dx^2 }=\frac{ -3y^2-9x^2 }{ y^3 }\]
put x=1,y=2
and get the solution.

Answer 3

So I would replace x with 1 and y with 2 and then simplify?