Answer to Question #180127 in Calculus for Osama

Question #180127

If𝑓(π‘₯,𝑦)=10π‘₯5βˆ’3π‘₯2𝑦6βˆ’4𝑦4.𝐹𝑖𝑛𝑑𝑓 (π‘₯,𝑦),𝑓 (π‘₯,𝑦), π‘₯π‘₯ π‘₯𝑦 𝑓 (1, βˆ’2), 𝑓 (1, 2)


1
Expert's answer
2021-04-29T17:13:50-0400

𝑓(π‘₯,𝑦)=10π‘₯5βˆ’3π‘₯2𝑦6βˆ’4𝑦4𝑓(π‘₯,𝑦)=10π‘₯^5βˆ’3π‘₯^2𝑦^6βˆ’4𝑦^4


Differentiate w.r.t x

f(x,y)x=50x4=6xy6f(x,y)_x=50x^4=6xy^6


Again differentiate w.r.t x-


f(x,y)xx=200x3βˆ’6y6f(x,y)_{xx}=200x^3-6y^6


f(1,2)xx=200(1)3βˆ’6(2)6=200βˆ’384=βˆ’184f(1,2)_{xx}=200(1)^3-6(2)^6=200-384=-184


f(1,βˆ’2)xx=200(1)3βˆ’6(βˆ’2)6=200βˆ’384=βˆ’184f(1,-2)_{xx}=200(1)^3-6(-2)^6=200-384=-184


Now differentiate w.r.t y-


f(x,y)y=βˆ’18x2y5βˆ’16y3f(x,y)_y=-18x^2y^5-16y^3


Again differentiate w.r.t. y-


f(x,y)yy=βˆ’90x2y4βˆ’48y2f(x,y)_{yy}=-90x^2y^4-48y^2


f(1,2)yy=βˆ’90(1)2(2)4βˆ’48(2)2=βˆ’1440βˆ’192=βˆ’1632f(1,2)_{yy}=-90(1)^2(2)^4-48(2)^2=-1440-192=-1632


f(1,βˆ’2)yy=βˆ’90(1)2(βˆ’2)4βˆ’48(βˆ’2)2=βˆ’1140βˆ’192=βˆ’1632f(1,-2)_{yy}=-90(1)^2(-2)^4-48(-2)^2=-1140-192=-1632


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