Math, asked by vishalchaudharu1405, 5 hours ago

Finde tlfx(x,y),fy(x,y),fx(1,3)and fy(_2,4) for the given function .if z=f (x,y)=3x3y2_x2y3+4x+9

Answers

Answered by pulakmath007
6

SOLUTION

GIVEN

 \sf{f(x, y)  = 3 {x}^{3} {y}^{2}  -  {x}^{2} {y}^{3}   + 4x + 9 }

TO DETERMINE

 \sf{f_x(x, y) \: , \:  f_y(x, y) \: , \:  f_x(1,3) \: , \: f_y(-2,4)}

EVALUATION

Here it is given that

 \sf{f(x, y)  = 3 {x}^{3} {y}^{2}  -  {x}^{2} {y}^{3}   + 4x + 9 }

Differentiating partially both sides with respect to x we get

 \sf{f_x (x, y)  = 9 {x}^{2} {y}^{2}  -  2{x}^{} {y}^{3}   + 4}

Putting x = 1 , y = 3 we get

 \sf{f_x (1, 3)  = 9  \times {1}^{2}  \times {3}^{2}  -  2 \times {1}^{} \times  {3}^{3}   + 4}

 \sf{ \implies \: f_x (1, 3)  = 81 - 54 + 4}

 \sf{ \implies \: f_x (1, 3)  = 31}

 \sf{f(x, y)  = 3 {x}^{3} {y}^{2}  -  {x}^{2} {y}^{3}   + 4x + 9 }

Differentiating partially with respect to y both sides we get

 \sf{f_y(x, y)  = 6 {x}^{3} {y}^{}  - 3 {x}^{2} {y}^{2}  }

Putting x = - 2 & y = 4 we get

 \sf{f_y( - 2, 4)  = 6  \times {( - 2)}^{3}  \times {4}^{}  - 3 \times  {( - 2)}^{2}  \times {4}^{2}  }

 \sf{ \implies \: f_y( - 2, 4)  =  - 192 - 192}

 \sf{ \implies \: f_y( - 2, 4)  =  - 384}

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