Question

Z=tg(xy^2) Найти:d^2z/dxdy,d^2z/dydx,dz

Answer 1

$z = \mathrm{tg}(xy^2)$

$\frac{\partial z}{\partial x} = \frac{1}{\cos^2(xy^2)}\cdot\frac{\partial (xy^2)}{\partial x} =$

$= \frac{y^2}{\cos^2(xy^2)}$

$\frac{\partial z}{\partial y} = \frac{1}{\cos^2(xy^2)}\cdot\frac{\partial (xy^2)}{\partial y} =$

$= \frac{2xy}{\cos^2(xy^2)}$

$\mathrm{d}z = \frac{\partial z}{\partial x}\cdot\mathrm{d}x + \frac{\partial z}{\partial y}\cdot\mathrm{d}y =$

$= \frac{y^2}{\cos^2(xy^2)}\cdot\mathrm{d}x + \frac{2xy}{\cos^2(xy^2)}\cdot\mathrm{d}y$

$\frac{\partial^2 z}{\partial x\partial y} = \frac{\partial }{\partial x} \frac{2xy}{\cos^2(xy^2)} =$

$= \frac{1}{\cos^4(xy^2)}\cdot\left( 2y\cdot\cos^2(xy^2) - 2xy\cdot 2\cos(xy^2)\cdot(-\sin(xy^2))\cdot y^2 \right) =$

$= \frac{1}{\cos^3(xy^2)}\cdot\left( 2y\cos(xy^2) + 4xy^3\sin(xy^2)\right) =$

$= \frac{ 2y\cdot(\cos(xy^2) + 2xy^2\sin(xy^2) )}{\cos^3(xy^2)}$

$\frac{\partial^2 z}{\partial y\partial x} = \frac{\partial }{\partial y} \frac{y^2}{\cos^2(xy^2)} =$

$= \frac{1}{\cos^4(xy^2)}\cdot\left( 2y\cos^2(xy^2) - y^2\cdot 2\cos(xy^2)\cdot(-\sin(xy^2))\cdot 2xy \right) =$

$= \frac{2y\cdot ( \cos(xy^2) + 2xy^2\sin(xy^2) )}{\cos^3(xy^2)}$

Answer 2

To find the partial derivatives and the total derivative of the given function $z = \tan(xy^2)$ with respect to $x$ and $y$, let's proceed step by step.

Given function: $z = \tan(xy^2)$

Partial Derivatives:

1. Partial derivative with respect to $x$ (first order): $\frac{\partial z}{\partial x} = \sec^2(xy^2) \cdot y^2$

2. Partial derivative with respect to $y$ (first order): $\frac{\partial z}{\partial y} = \sec^2(xy^2) \cdot 2xy$

Second-Order Partial Derivatives:

3. Second partial derivative with respect to $x$ twice: $\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)$ $\frac{\partial^2 z}{\partial x^2} = 2y^2 \cdot \sec^2(xy^2) \cdot \tan(xy^2) + \sec^4(xy^2) \cdot y^4$

4. Second partial derivative with respect to $y$ twice: $\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)$ $\frac{\partial^2 z}{\partial y^2} = 2x \cdot \sec^2(xy^2) + 2\cdot 2x^2y^2 \cdot \sec^2(xy^2) \cdot 2xy + 4x^2 \cdot \sec^2(xy^2) \cdot y^2$

Mixed Partial Derivatives:

5. Mixed partial derivative with respect to $x$ and $y$ (first $x$, then $y$): $\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)$ $\frac{\partial^2 z}{\partial x \partial y} = 4xy \cdot \sec^2(xy^2) + 2y \cdot \sec^2(xy^2) \cdot 2xy + 2 \cdot 2y^3 \cdot \sec^2(xy^2) \cdot y^2$

6. Mixed partial derivative with respect to $y$ and $x$ (first $y$, then $x$): $\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)$