Question

Tìm vi phân toàn phần của các hàm số sau:
ln(x+√(x^2+y^2 ) ) ln(sin(y/x))

Answer 1

Let f = ln(x + √(x ² + y ²)) ln(sin(y/x)).

Then the total differential is

`\mathrm df = (\mathrm d\left(x+√(x^2+y^2)\right))/(x+√(x^2+y^2))\ln\left(\sin\left(\frac yx\right)\right) + \ln\left(x+√(x^2+y^2)\right)(\mathrm d\left(\sin\left(\frac yx\right)\right))/(\sin\left(\frac yx\right))`

`\mathrm df = (\mathrm dx + (\mathrm d(x^2+y^2))/(√(x^2+y^2)))/(x+√(x^2+y^2))\ln\left(\sin\left(\frac yx\right)\right) + \ln\left(x+√(x^2+y^2)\right)(\cos\left(\frac yx\right)\,\mathrm d\left(\frac yx\right))/(\sin\left(\frac yx\right))`

`\mathrm df = (\mathrm dx + (2x\,\mathrm dx+2y\,\mathrm dy)/(√(x^2+y^2)))/(x+√(x^2+y^2)\right)\ln\left(\sin\left(\frac yx\right)\right) + \ln\left(x+√(x^2+y^2))\right)(\cos\left(\frac yx\right)(x\,\mathrm dy-y\,\mathrm dx)/(x^2))/(\sin\left(\frac yx\right))`

`\mathrm df = \frac{\left(2x+√(x^2+y^2)\right)\,\mathrm dx +2y\,\mathrm dy}{x√(x^2+y^2)+x^2+y^2\right)\ln\left(\sin\left(\frac yx\right)\right) \\\\ \indent + \frac1{x^2}\cot\left(\frac yx\right)\ln\left(x+√(x^2+y^2)}\right)(x\,\mathrm dy-y\,\mathrm dx)`

`\mathrm df = \left(\left((2x+√(x^2+y^2))/(x√(x^2+y^2)+x^2+y^2)\right)\ln\left(\sin\left(\frac yx\right)\right) - \frac y{x^2}\cot\left(\frac yx\right)\ln\left(x+√(x^2+y^2)\right)\right)\,\mathrm dx \\\\ \indent + \left((2y)/(x√(x^2+y^2)+x^2+y^2)\ln\left(\sin\left(\frac yx\right)\right)+\frac1x\cot\left(\frac yx\right)\ln\left(x+√(x^2+y^2)\right)\right)\,\mathrm dy`