Question

find the first order partial differentiation of cos-1 (x/y)​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{f=cos^{-1}\left(\dfrac{x}{y}\right)}$

$\underline{\textbf{To find:}}$

$\textsf{First order partial derivatives of f}$

$\underline{\textbf{Solution:}}$

$\mathsf{f=cos^{-1}\left(\dfrac{x}{y}\right)}$

$\implies\mathsf{\dfrac{{\partial}f}{{\partial}x}=\dfrac{-1}{1+\left(\dfrac{x}{y}\right)^2}{\times}\dfrac{1}{y}}$

$\implies\mathsf{\dfrac{{\partial}f}{{\partial}x}=\dfrac{-y^2}{y^2+x^2}{\times}\dfrac{1}{y}}$

$\implies\boxed{\mathsf{\dfrac{{\partial}f}{{\partial}x}=\dfrac{-y}{x^2+y^2}}}$

$\mathsf{f=cos^{-1}\left(\dfrac{x}{y}\right)}$

$\implies\mathsf{\dfrac{{\partial}f}{{\partial}y}=\dfrac{-1}{1+\left(\dfrac{x}{y}\right)^2}{\times}\dfrac{-x}{y^2}}$

$\implies\mathsf{\dfrac{{\partial}f}{{\partial}y}=\dfrac{-y^2}{y^2+x^2}{\times}\dfrac{-x}{y^2}}$

$\implies\boxed{\mathsf{\dfrac{{\partial}f}{{\partial}y}=\dfrac{x}{x^2+y^2}}}$

$\underline{\textbf{Find more:}}$

Find the second order partial derivatives of F(x.y) = (3x + 2y).4​

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If u=x³y³/x³+y³,prove that x du/dx + y du/dy=3u​

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Answer 2

given,

$f =cos-1 (x/y)\\Find$

First  order partial derivatives of f

$f = cos-1 (x/y)\\df/dx =\frac{-1}{1+(x/y)^2} *\frac{1}{y} \\df/dx =\frac{-y^2}{y^2+x^2} *\frac{1}{y} \\df/dx = \frac{-y}{x^2 +y^2}$