Penyelesaian - Derivative

Applying the product rule of derivatives.

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

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

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

Computing the derivative of a power function.

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

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

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

The derivative of a constant value is always zero.

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

Multiplying a number by zero always results in zero.

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

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

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