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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).

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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.

This problem has been solved

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