Suppose thatF (x, y) = g(u(x, y), v(x, y))with u(x, y) = x2 and v(x, y) = h(x, y) − x. Furthermore suppose thath(1, 1) = 2, hx(1, 1) = 3, hy(1, 1) = −2, g(1, 1) = 6, gu(1, 1) = −4 andgv(1, 1) = 5. Determine the partial derivatives Fx(1, 1) and Fy(1, 1).
Question
Suppose thatF (x, y) = g(u(x, y), v(x, y))with u(x, y) = x2 and v(x, y) = h(x, y) − x. Furthermore suppose thath(1, 1) = 2, hx(1, 1) = 3, hy(1, 1) = −2, g(1, 1) = 6, gu(1, 1) = −4 andgv(1, 1) = 5. Determine the partial derivatives Fx(1, 1) and Fy(1, 1).
Solution
To find the partial derivatives Fx(1,1) and Fy(1,1), we need to use the chain rule for partial derivatives.
The chain rule for two variables states that:
Fx = Fu * ux + Fv * vx Fy = Fu * uy + Fv * vy
Given that u(x, y) = x^2 and v(x, y) = h(x, y) - x, we can find that:
ux = 2x uy = 0 vx = hx - 1 vy = hy
Substituting x = 1, y = 1, we get:
ux(1,1) = 2 uy(1,1) = 0 vx(1,1) = hx(1,1) - 1 = 3 - 1 = 2 vy(1,1) = hy(1,1) = -2
Now, we can find Fx(1,1) and Fy(1,1) using the chain rule:
Fx(1,1) = Fu(1,1) * ux(1,1) + Fv(1,1) * vx(1,1) = gu(1,1) * ux(1,1) + gv(1,1) * vx(1,1) = -4 * 2 + 5 * 2 = -8 + 10 = 2
Fy(1,1) = Fu(1,1) * uy(1,1) + Fv(1,1) * vy(1,1) = gu(1,1) * uy(1,1) + gv(1,1) * vy(1,1) = -4 * 0 + 5 * -2 = -10
So, Fx(1,1) = 2 and Fy(1,1) = -10.
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