Question

Partial Differential co-efficient of U = sin(ax + by + cz).

Answer 1

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

$\mathsf{U=sin(ax+by+cz)}$

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

$\textsf{Partial differential coefficient of U}$

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

$\mathsf{Consider,}$

$\mathsf{U=sin(ax+by+cz)}$

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

$\mathsf{\dfrac{\partial{U}}{\partial{x}}=cos(ax+by+cz).\dfrac{\partial(ax+by+cz)}{\partial{x}}}$

$\mathsf{\dfrac{\partial{U}}{\partial{x}}=cos(ax+by+cz).a}$

$\implies\boxed{\mathsf{\dfrac{\partial{U}}{\partial{x}}=a\,cos(ax+by+cz)}}$

$\mathsf{U=sin(ax+by+cz)}$

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

$\mathsf{\dfrac{\partial{U}}{\partial{y}}=cos(ax+by+cz).\dfrac{\partial(ax+by+cz)}{\partial{y}}}$

$\mathsf{\dfrac{\partial{U}}{\partial{y}}=cos(ax+by+cz).b}$

$\implies\boxed{\mathsf{\dfrac{\partial{U}}{\partial{y}}=b\,cos(ax+by+cz)}}$

$\mathsf{U=sin(ax+by+cz)}$

$\textsf{Differentiate partially with respect to 'z'}$

$\mathsf{\dfrac{\partial{U}}{\partial{z}}=cos(ax+by+cz).\dfrac{\partial(ax+by+cz)}{\partial{z}}}$

$\mathsf{\dfrac{\partial{U}}{\partial{z}}=cos(ax+by+cz).c}$

$\implies\boxed{\mathsf{\dfrac{\partial{U}}{\partial{z}}=c\,cos(ax+by+cz)}}$

#SPJ3