Answer to Question #108363 in Calculus for Nimra

Question

Find first-order partial derivatives of F(x)= cosx²y²z²/Arc sinxy²

Answer 1

Consider the function $f(x,y,z)=\frac{\cos(x^2y^2z^2)}{\arcsin(xy^2)}$

Differentiate the function partially with respect to $x$ as,

$$f_x(x,y,z)=\frac{\partial}{\partial x}(\frac{\cos(x^2y^2z^2)}{\arcsin(xy^2)})$$

$$=\frac{1}{\arcsin^2(xy^2)}(\arcsin(xy^2)\frac{\partial}{\partial x}(\cos(x^2y^2z^2))-(\cos(x^2y^2z^2))\frac{\partial}{\partial x}(\arcsin(xy^2)))$$

$$=\frac{-2xy^2z^2\sin(x^2y^2z^2)\arcsin(xy^2)-\frac{y^2\cos(x^2y^2z^2)}{\sqrt{1-x^2y^4}}}{\arcsin^2(xy^2)}$$

$$=\frac{-2xy^2z^2\sin(x^2y^2z^2)\arcsin(xy^2)\sqrt{1-x^2y^4}-y^2\cos(x^2y^2z^2)}{\arcsin^2(xy^2)\sqrt{1-x^2y^4}}$$

Differentiate the function partially with respect to $y$ as,

$$f_y(x,y,z)=\frac{\partial}{\partial y}(\frac{\cos(x^2y^2z^2)}{\arcsin(xy^2)})$$

$$=\frac{1}{\arcsin^2(xy^2)}(\arcsin(xy^2)\frac{\partial}{\partial y}(\cos(x^2y^2z^2))-(\cos(x^2y^2z^2))\frac{\partial}{\partial y}(\arcsin(xy^2)))$$

$$=\frac{-2x^2yz^2\sin(x^2y^2z^2)\arcsin(xy^2)-\frac{2xy\cos(x^2y^2z^2)}{\sqrt{1-x^2y^4}}}{\arcsin^2(xy^2)}$$

$$=\frac{-2x^2yz^2\sin(x^2y^2z^2)\arcsin(xy^2)\sqrt{1-x^2y^4}-2xy\cos(x^2y^2z^2)}{\arcsin^2(xy^2)\sqrt{1-x^2y^4}}$$

Differentiate the function with respect to $z$ as,

$$f_z(x,y,z)=\frac{\partial}{\partial z}(\frac{\cos(x^2y^2z^2)}{\arcsin(xy^2)})$$

$$=\frac{1}{\arcsin(xy^2)}\frac{\partial}{\partial z}(\cos(x^2y^2z^2))$$

$$=\frac{-2x^2y^2z\sin(x^2y^2z^2)}{\arcsin(xy^2)}$$

Therefore, the partial derivatives of the multi variable function are:

$$f_x(x,y,z)=\frac{-2xy^2z^2\sin(x^2y^2z^2)\arcsin(xy^2)\sqrt{1-x^2y^4}-y^2\cos(x^2y^2z^2)}{\arcsin^2(xy^2)\sqrt{1-x^2y^4}}$$

$$f_y(x,y,z=\frac{-2x^2yz^2\sin(x^2y^2z^2)\arcsin(xy^2)\sqrt{1-x^2y^4}-2xy\cos(x^2y^2z^2)}{\arcsin^2(xy^2)\sqrt{1-x^2y^4}}$$

$$f_z(x,y,z)=\frac{-2x^2y^2z\sin(x^2y^2z^2)}{\arcsin(xy^2)}$$