Question : If u=sin^(-1){(x+2y+3z)/(x^8+y^8+z^8)} then find the value of xdf/dx+ydf/dy+zdf/dz Solution : If \[u=sin^{-1}\left(\frac{x+2y+3z}{x^8+y^8+z^8}\right)\] then=>\[ sin{\left(u\right)}=\frac{x+2y+3z}{x^8+y^8+z^8}\]Differentiate with respect to x\[cos{\left(u\right)}\frac{\partial u}{\partial x}=\frac{\left(x^8+y^8+z^8\right).1-\left(x+2y+3z\right).\left(8x^7\right)}{\left(x^8+y^8+z^8\right)^2}……….(1)\]Similarly differentiating with respect to y and z \[cos{\left(u\right)}\frac{\partial u}{\partial y}=\frac{\left(x^8+y^8+z^8\right).2-\left(x+2y+3z\right).\left(8y^7\right)}{\left(x^8+y^8+z^8\right)^2}……….(2)\]\[cos{\left(u\right)}\frac{\partial u}{\partial z}=\frac{\left(x^8+y^8+z^8\right).3-\left(x+2y+3z\right).\left(8z^7\right)}{\left(x^3+y^3+z^3\right)^2}……….(2)\]Multiplying eqns (1) ,(2) and (3) by…
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