text: This is a directional derivative and you have to have a direction to the derivative. Any way let the direction be (th). Then\n\nw = f(y,x)\nand\ny = f(s,t) \nand\nx = f(s,t)\n\nNote: dw/dt means the total derivative of w in the direction of (th) and it’s equals to\n\ndw/dt = (Pw/Pt) sin(th) + (Pw/Ps) cos(th) ……. where P is partial derivative\n\nAlso,\n\nPw/Pt = Pw/Py * Py/Pt + Pw/Px * Px/Pt……. where P is partial derivative\n\nFor\n\nw= (siny) exp(x) …. And x = s t …… and y = s + t\n\nthen\n\nPw/Pt = (cos y) exp(x) * 1 + (sin y) exp(x) *s\n\nAnd\n\nPw/Ps = Pw/Py * Py/Ps + Pw/Px * Px/Ps\n\nPw/Ps = (cosy) exp(x) * 1 + (siny) exp(x) *t\n\nfor dw/dt\n\ndw/dt = [ (cos y) exp(x) * 1 + (sin y) exp(x) *s] sin(th) + [(cosy) exp(x) * 1 + (siny) exp(x) *t] cos(th)\n\nNow\n\nd^2w/dsdt = d/ds[dw/dt]\n\nthen\n\nd^2w/dsdt = d/ds{[ (cos y) exp(x) * 1 + (sin y) exp(x) *s] sin(th) + [(cosy) exp(x) * 1 + (siny) exp(x) *t] cos(th)}\n\nd^2w/dsdt = [(siny) exp(x)] sin(th) ……….. eq(1)\n\nfor s =1 and t = 1 --->  x  = 1   and  y = 2\n\nSubstitute for x and y in eq.1\n\nd^2w/dsdt = [sin(2) exp(1)] sin(th) \n\nand that is the second mix directional derivative of w at ( s =1 and t =1 ) in the direction of (th)