Answer:
[tex]f_x(x,y)=6x^{5}+9y^5[/tex]
[tex]f_y(x,y)=45xy^{4}[/tex]
Step-by-step explanation:
Given : Function [tex]f(x,y)=x^6+9xy^5[/tex]
To find : The first partial derivatives of the function fx(x, y) and fy(x, y) ?
Solution :
Function [tex]f(x,y)=x^6+9xy^5[/tex]
Partial derivative w.r.t to x means y treat as constant,
[tex]f_x(x,y)=6x^{6-1}+9\times 1x^{1-1}y^5[/tex]
[tex]f_x(x,y)=6x^{5}+9x^{0}y^5[/tex]
[tex]f_x(x,y)=6x^{5}+9y^5[/tex]
Partial derivative w.r.t to y means x treat as constant,
[tex]f_y(x,y)=0+9\times 5xy^{5-1}[/tex] (Derivative of constant is zero)
[tex]f_y(x,y)=45xy^{4}[/tex]
