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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.
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Do you think you can help? @surjithayer
\[3x^2+y^2=7\] 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.
So I would replace x with 1 and y with 2 and then simplify?
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