Question

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

Answer 1

$\textbf{Given:}$

$\mathsf{F(x,y)=(3x+2y)^4}$

$\textbf{To find:}$

$\textsf{Second order partial derivatives of F}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{F(x,y)=(3x+2y)^4}$

$\textsf{Differentiate partially with respect to x}$

$\mathsf{\dfrac{\partial\,F}{\partial\,x}=4(3x+2y)^3(3)=12(3x+2y)^3}$

$\textsf{Differentiate again partially with respect to x}$

$\mathsf{\dfrac{{\partial}^2F}{\partial\,x^2}=36(3x+2y)^2(3)=108(3x+2y)^2}$

$\implies\boxed{\mathsf{\dfrac{{\partial}^2F}{\partial\,x^2}=108(3x+2y)^2}}$

$\mathsf{F(x,y)=(3x+2y)^4}$

$\textsf{Differentiate partially with respect to y}$

$\mathsf{\dfrac{\partial\,F}{\partial\,y}=4(3x+2y)^3(2)=8(3x+2y)^3}$

$\textsf{Differentiate again partially with respect to y}$

$\mathsf{\dfrac{{\partial}^2F}{\partial\,y^2}=24(3x+2y)^2(2)=48(3x+2y)^2}$

$\implies\boxed{\mathsf{\dfrac{{\partial}^2F}{\partial\,y^2}=48(3x+2y)^2}}$

$\textbf{Find more:}$

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

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