Question

Find the first partial derivatives of the function. f(x, y) = x6 + 9xy5 fx(x, y) = _________ fy(x, y) = _____________

Answer 1

Answer:

$f_x(x,y)=6x^{5}+9y^5$

$f_y(x,y)=45xy^{4}$

Step-by-step explanation:

Given : Function $f(x,y)=x^6+9xy^5$

To find : The first partial derivatives of the function fx(x, y) and fy(x, y) ?

Solution :

Function $f(x,y)=x^6+9xy^5$

Partial derivative w.r.t to x means y treat as constant,

$f_x(x,y)=6x^{6-1}+9\times 1x^{1-1}y^5$

$f_x(x,y)=6x^{5}+9x^{0}y^5$

$f_x(x,y)=6x^{5}+9y^5$

Partial derivative w.r.t to y means x treat as constant,

$f_y(x,y)=0+9\times 5xy^{5-1}$ (Derivative of constant is zero)

$f_y(x,y)=45xy^{4}$