Řešení - Derivative

Applying the sum rule of derivatives.

$$\frac{d}{dx}[x{y}^2+x^{2}y]=\frac{d}{dx}\left[x{y}^2\right]+\frac{d}{dx}\left[x^{2}y\right]$$

Applying the product rule of derivatives.

$$\frac{d}{dx}\left[x{y}^2\right]+\frac{d}{dx}\left[x^{2}y\right]=\left(\frac{d}{dx}\left[x\right]\times y^2+x\times \frac{d}{dx}\left[y^2\right]\right)+\frac{d}{dx}\left[x^{2}y\right]$$

Applying the product rule of derivatives.

$$\left(\frac{d}{dx}\left[x\right]\times y^2+x\times \frac{d}{dx}\left[y^2\right]\right)+\frac{d}{dx}\left[x^{2}y\right]=\left(\frac{d}{dx}\left[x\right]\times y^2+x\times \frac{d}{dx}\left[y^2\right]\right)+\left(\frac{d}{dx}\left[x^2\right]\times y+x^2\times \frac{d}{dx}\left[y\right]\right)$$

The derivative of a variable with respect to itself is always equal to one.

$$\left(\frac{d}{dx}\left[x\right]\times y^2+x\times \frac{d}{dx}\left[y^2\right]\right)+\left(\frac{d}{dx}\left[x^2\right]\times y+x^2\times \frac{d}{dx}\left[y\right]\right)=\left(1y^2+x\times \frac{d}{dx}\left[y^2\right]\right)+\left(\frac{d}{dx}\left[x^2\right]\times y+x^2\times \frac{d}{dx}\left[y\right]\right)$$

Computing the derivative of a power function.

$$\left(1y^2+x\times \frac{d}{dx}\left[y^2\right]\right)+\left(\frac{d}{dx}\left[x^2\right]\times y+x^2\times \frac{d}{dx}\left[y\right]\right)=\left(1y^2+x\times \left(y^2\left(\frac{d}{dx}\left[2\right]\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(\frac{d}{dx}\left[x^2\right]\times y+x^2\times \frac{d}{dx}\left[y\right]\right)$$

Computing the derivative of x raised to the power of n.

$$\left(1y^2+x\times \left(y^2\left(\frac{d}{dx}\left[2\right]\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(\frac{d}{dx}\left[x^2\right]\times y+x^2\times \frac{d}{dx}\left[y\right]\right)=\left(1y^2+x\times \left(y^2\left(\frac{d}{dx}\left[2\right]\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(\left(2x^{2-1}\right)\times y+x^2\times \frac{d}{dx}\left[y\right]\right)$$

Subtracting one from a number.

$$\left(1y^2+x\times \left(y^2\left(\frac{d}{dx}\left[2\right]\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(\left(2x^{2-1}\right)\times y+x^2\times \frac{d}{dx}\left[y\right]\right)=\left(1y^2+x\times \left(y^2\left(\frac{d}{dx}\left[2\right]\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(\left(2x^1\right)\times y+x^2\times \frac{d}{dx}\left[y\right]\right)$$

Any number raised to the power of one equals the number itself.

$$\left(1y^2+x\times \left(y^2\left(\frac{d}{dx}\left[2\right]\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(\left(2x^1\right)\times y+x^2\times \frac{d}{dx}\left[y\right]\right)=\left(1y^2+x\times \left(y^2\left(\frac{d}{dx}\left[2\right]\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(\left(2x\right)\times y+x^2\times \frac{d}{dx}\left[y\right]\right)$$

Multiplying a number by one, which does not change its value.

$$\left(1y^2+x\times \left(y^2\left(\frac{d}{dx}\left[2\right]\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(\left(2x\right)\times y+x^2\times \frac{d}{dx}\left[y\right]\right)=\left(y^2+x\times \left(y^2\left(\frac{d}{dx}\left[2\right]\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(\left(2x\right)\times y+x^2\times \frac{d}{dx}\left[y\right]\right)$$

The derivative of a constant value is always zero.

$$\left(y^2+x\times \left(y^2\left(\frac{d}{dx}\left[2\right]\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(\left(2x\right)\times y+x^2\times \frac{d}{dx}\left[y\right]\right)=\left(y^2+x\times \left(y^2\left(0\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(\left(2x\right)\times y+x^2\times \frac{d}{dx}\left[y\right]\right)$$

Multiplying a number by zero always results in zero.

$$\left(y^2+x\times \left(y^2\left(0\times \ln \left(y\right)+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(2x\times y+x^2\times \frac{d}{dx}\left[y\right]\right)=\left(y^2+x\times \left(y^2\left(0+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(2x\times y+x^2\times \frac{d}{dx}\left[y\right]\right)$$

Simplifying the arithmetic expressions.

$$\left(y^2+x\times \left(y^2\left(0+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(2x\times y+x^2\times \frac{d}{dx}\left[y\right]\right)=\left(y^2+x\times \left(y^2\left(0+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(2xy+x^2\times \frac{d}{dx}\left[y\right]\right)$$

Adding zero to a number, which does not change its value.

$$\left(y^2+x\times \left(y^2\left(0+\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(2xy+x^2\times \frac{d}{dx}\left[y\right]\right)=\left(y^2+x\times \left(y^2\times \left(\frac{2}{y}\times \frac{d}{dx}\left[y\right]\right)\right)\right)+\left(2xy+x^2\times \frac{d}{dx}\left[y\right]\right)$$