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Find the directional derivative of f(x, y)?

Find the directional derivative of f(x, y)=x^2*y+4y^2 at (2, 1) for u=<1/2, sqrt(3)/2>.

Answer: 2+6sqrt(3)

<2xy, 8y> at (2, 1)=<4, 8>

<4, 8>*<1/2, sqrt(3)/2>=2+4sqrt(3)

This doesn't match the answer so can anyone check my work and correct me if I'm wrong? Show your work.

Comments

x^2*y means (x^2)(y), not x^(2y). And how should I find the derivative to this with respect to y?

You need to treat x as a constant not a variable. For the x2y you must think of x2 as a scalar not a variable. So follow the simple rule of derivatives d/dx[cu] = cu' and you get x2. It is as if you are taking the derivative of a single variable function and you have f(x) = 6x. The derivative in this case would be 6 and just like that case the partial derivative with respect to y is x2 + 8y.

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1 Answer

the derivative with respect to y is 4x^2 + 8y. An easy mistake to make. Retry the dot product now and I'm sure you will get the answer.

Comments

The derivative wrt to y = x^2  +  8y.  Sun, it would help considerably if you would use parenthesis and spacing to advantage.  For instance, x^2*y+4y^2, the first term may be considered x^(2y) and then the second is 4y^2.  Reverse engineering from your given answer gives it as (x^2)y + 4y^2.  I'm sure that I am not the only one who loses interest after working , only to find the error is due to the presentation.

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