Question

$\rm Find \: f_x \: and \: f_y \: where \: \rm f(x,y)=cos(x^2y)+y^3$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x,y)=cos(x^2y)+y^3$

can be rewritten as

$\rm \: f=cos(x^2y)+y^3$

On differentiating partially w. r. t. x, we get

$\rm \: \dfrac{\partial }{\partial x}f=\dfrac{\partial }{\partial x}[cos(x^2y)+y^3 ]$

$\rm \: f_x=\dfrac{\partial }{\partial x}cos(x^2y)+\dfrac{\partial }{\partial x}y^3$

$\rm \: f_x= - sin( {x}^{2}y) \dfrac{\partial }{\partial x}(x^2y) + 0$

$\rm \: f_x= - y \: sin( {x}^{2}y) \dfrac{\partial }{\partial x}(x^2)$

$\rm \: f_x= - y \: sin( {x}^{2}y) (2x)$

$\rm \: f_x= -2x y \: sin( {x}^{2}y)$

So,

$\rm\implies \:\boxed{ \rm{ \:\rm \: f_x= -2x y \: sin( {x}^{2}y) \: }}$

Now, Consider again

$\rm \:f =cos(x^2y)+y^3$

On differentiating partially w. r. t. y, we get

$\rm \:\dfrac{\partial }{\partial y}f =\dfrac{\partial }{\partial y}[cos(x^2y)+y^3]$

$\rm \:f_y =\dfrac{\partial }{\partial y}cos(x^2y)+\dfrac{\partial }{\partial y}y^3$

$\rm \:f_y = - sin( {x}^{2}y)\dfrac{\partial }{\partial y}(x^2y) + 3y^2$

$\rm \:f_y = - {x}^{2} sin( {x}^{2}y)\dfrac{\partial }{\partial y}(y) + 3y^2$

$\rm \:f_y = - {x}^{2} sin( {x}^{2}y) + 3y^2$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \:f_y = - {x}^{2} sin( {x}^{2}y) + 3y^2 \: }}$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:\dfrac{\partial }{\partial x}cosx \: = \: - \: sinx \: }}$

$\boxed{ \rm{ \:\dfrac{\partial }{\partial x} {x}^{n} \: = \: {nx}^{n - 1} \: }}$

$\boxed{ \rm{ \:\dfrac{\partial }{\partial x}k \: = \: 0 \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$