Penyelesaian - Derivative

Applying the product rule of derivatives.

$$\frac{d}{dx}[4\times \frac{{\left(xy\right)}^1}{2}]=\frac{d}{dx}\left[4\right]\times \frac{{\left(xy\right)}^1}{2}+4\times \frac{d}{dx}\left[\frac{{\left(xy\right)}^1}{2}\right]$$

The derivative of a constant value is always zero.

$$\frac{d}{dx}\left[4\right]\times \frac{{\left(xy\right)}^1}{2}+4\times \frac{d}{dx}\left[\frac{{\left(xy\right)}^1}{2}\right]=0\times \frac{{\left(xy\right)}^1}{2}+4\times \frac{d}{dx}\left[\frac{{\left(xy\right)}^1}{2}\right]$$

Multiplying a number by zero always results in zero.

$$0\times \frac{{\left(xy\right)}^1}{2}+4\times \frac{d}{dx}\left[\frac{{\left(xy\right)}^1}{2}\right]=0+4\times \frac{d}{dx}\left[\frac{{\left(xy\right)}^1}{2}\right]$$

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

$$0+4\times \frac{d}{dx}\left[\frac{{\left(xy\right)}^1}{2}\right]=4\times \frac{d}{dx}\left[\frac{{\left(xy\right)}^1}{2}\right]$$

Computing the derivative of a fraction.

$$4\times \frac{d}{dx}\left[\frac{{\left(xy\right)}^1}{2}\right]=4\times \frac{\frac{d}{dx}\left[{\left(xy\right)}^1\right]\times 2-{\left(xy\right)}^1\times \frac{d}{dx}\left[2\right]}{2^2}$$

Computing the derivative of a power function.

$$4\times \frac{\frac{d}{dx}\left[{\left(xy\right)}^1\right]\times 2-{\left(xy\right)}^1\times \frac{d}{dx}\left[2\right]}{2^2}=4\times \frac{\left({\left(xy\right)}^1\times \left(\frac{d}{dx}\left[1\right]\times \ln \left(xy\right)+\frac{1}{xy}\times \frac{d}{dx}\left[xy\right]\right)\right)\times 2-{\left(xy\right)}^1\times \frac{d}{dx}\left[2\right]}{2^2}$$

The derivative of a constant value is always zero.

$$4\times \frac{\left({\left(xy\right)}^1\times \left(\frac{d}{dx}\left[1\right]\times \ln \left(xy\right)+\frac{1}{xy}\times \frac{d}{dx}\left[xy\right]\right)\right)\times 2-{\left(xy\right)}^1\times \frac{d}{dx}\left[2\right]}{2^2}=4\times \frac{\left({\left(xy\right)}^1\times \left(\frac{d}{dx}\left[1\right]\times \ln \left(xy\right)+\frac{1}{xy}\times \frac{d}{dx}\left[xy\right]\right)\right)\times 2-{\left(xy\right)}^1\times 0}{2^2}$$

Simplifying the arithmetic expressions.

$$4\times \frac{\left({\left(xy\right)}^1\times \left(\frac{d}{dx}\left[1\right]\times \ln \left(xy\right)+\frac{1}{xy}\times \frac{d}{dx}\left[xy\right]\right)\right)\times 2-{\left(xy\right)}^1\times 0}{2^2}=4\times \frac{\left({\left(xy\right)}^1\times \left(\frac{d}{dx}\left[1\right]\times \ln \left(xy\right)+\frac{1}{xy}\times \frac{d}{dx}\left[xy\right]\right)\right)\times 2-{\left(xy\right)}^1\times 0}{4}$$

The derivative of a constant value is always zero.

$$4\times \frac{\left({\left(xy\right)}^1\times \left(\frac{d}{dx}\left[1\right]\times \ln \left(xy\right)+\frac{1}{xy}\times \frac{d}{dx}\left[xy\right]\right)\right)\times 2-{\left(xy\right)}^1\times 0}{4}=4\times \frac{\left({\left(xy\right)}^1\times \left(0\times \ln \left(xy\right)+\frac{1}{xy}\times \frac{d}{dx}\left[xy\right]\right)\right)\times 2-{\left(xy\right)}^1\times 0}{4}$$

Applying the product rule of derivatives.

$$4\times \frac{\left({\left(xy\right)}^1\times \left(0\times \ln \left(xy\right)+\frac{1}{xy}\times \frac{d}{dx}\left[xy\right]\right)\right)\times 2-{\left(xy\right)}^1\times 0}{4}=4\times \frac{\left({\left(xy\right)}^1\times \left(0\times \ln \left(xy\right)+\frac{1}{xy}\times \left(\frac{d}{dx}\left[x\right]\times y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2-{\left(xy\right)}^1\times 0}{4}$$

Multiplying a number by zero always results in zero.

$$4\times \frac{\left({\left(xy\right)}^1\times \left(0\times \ln \left(xy\right)+\frac{1}{xy}\times \left(\frac{d}{dx}\left[x\right]\times y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2-{\left(xy\right)}^1\times 0}{4}=4\times \frac{\left({\left(xy\right)}^1\times \left(0\times \ln \left(xy\right)+\frac{1}{xy}\times \left(\frac{d}{dx}\left[x\right]\times y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2+0}{4}$$

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

$$4\times \frac{\left({\left(xy\right)}^1\times \left(0\times \ln \left(xy\right)+\frac{1}{xy}\times \left(\frac{d}{dx}\left[x\right]\times y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2+0}{4}=4\times \frac{\left({\left(xy\right)}^1\times \left(0\times \ln \left(xy\right)+\frac{1}{xy}\times \left(\frac{d}{dx}\left[x\right]\times y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2}{4}$$

Simplifying the arithmetic expressions.

$$4\times \frac{\left({\left(xy\right)}^1\times \left(0\times \ln \left(xy\right)+\frac{1}{xy}\times \left(\frac{d}{dx}\left[x\right]\times y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2}{4}=4\times \frac{\left(xy\times \left(0\times \ln \left(xy\right)+\frac{1}{xy}\times \left(\frac{d}{dx}\left[x\right]\times y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2}{4}$$

Multiplying a number by zero always results in zero.

$$4\times \frac{\left(xy\times \left(0\times \ln \left(xy\right)+\frac{1}{xy}\times \left(\frac{d}{dx}\left[x\right]\times y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2}{4}=4\times \frac{\left(xy\times \left(0+\frac{1}{xy}\times \left(\frac{d}{dx}\left[x\right]\times y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2}{4}$$

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

$$4\times \frac{\left(xy\times \left(0+\frac{1}{xy}\times \left(\frac{d}{dx}\left[x\right]\times y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2}{4}=4\times \frac{\left(xy\times \left(0+\frac{1}{xy}\times \left(1y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2}{4}$$

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

$$4\times \frac{\left(xy\times \left(0+\frac{1}{xy}\times \left(1y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2}{4}=4\times \frac{\left(xy\times \left(\frac{1}{xy}\times \left(1y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2}{4}$$

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

$$4\times \frac{\left(xy\times \left(\frac{1}{xy}\times \left(1y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2}{4}=4\times \frac{\left(xy\times \left(\frac{1}{xy}\times \left(y+x\times \frac{d}{dx}\left[y\right]\right)\right)\right)\times 2}{4}$$