I need some help finding the second derivative with respect to x of f(x, y) = sin(mx + ny)2
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$\displaystyle \frac{\partial}{\partial x}sin^2(mx+ny)= 2msin(mx+ny)cos(mx+ny)$ $\displaystyle \frac{\partial}{\partial x}2msin(mx+ny)cos(mx+ny)= 2m^2cos^2(mx+ny)-2m^2sin^2(mx+ny)$
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