해결방법 - 미분

곱셈에 대한 도함수를 확장합니다.

$$\frac{d}{dx}[e^c\times osx]=\frac{d}{dx}\left[e^c\right]\times osx+e^c\times \frac{d}{dx}\left[o\right]\times sx+e^c\times o\times \frac{d}{dx}\left[s\right]\times x+e^c\times os\times \frac{d}{dx}\left[x\right]$$

곱셈에 대한 도함수를 확장합니다.

$$\frac{d}{dx}[e^c\times osx]=\frac{d}{dx}\left[e^c\right]\times osx+e^c\times \frac{d}{dx}\left[o\right]\times sx+e^c\times o\times \frac{d}{dx}\left[s\right]\times x+e^c\times os\times \frac{d}{dx}\left[x\right]$$

곱셈은 다르게 그룹화해도 결과가 같음.

$$\frac{d}{dx}[e^c\times osx]=\frac{d}{dx}[e^c\times \left(osx\right)]$$

도함수의 곱의 규칙을 적용합니다.

$$\frac{d}{dx}[e^c\times \left(osx\right)]=\frac{d}{dx}\left[e^c\right]\times \left(osx\right)+e^c\times \frac{d}{dx}\left[osx\right]$$

곱셈에 대한 도함수를 확장합니다.

$$\frac{d}{dx}[e^c\times osx]=\frac{d}{dx}\left[e^c\right]\times osx+e^c\times \frac{d}{dx}\left[o\right]\times sx+e^c\times o\times \frac{d}{dx}\left[s\right]\times x+e^c\times os\times \frac{d}{dx}\left[x\right]$$

곱셈에 대한 도함수를 확장합니다.

$$\frac{d}{dx}\left[e^c\right]\times \left(osx\right)+e^c\times \frac{d}{dx}\left[osx\right]=\frac{d}{dx}\left[e^c\right]\times \left(osx\right)+e^c\times \left(\frac{d}{dx}\left[o\right]\times sx+o\times \frac{d}{dx}\left[s\right]\times x+os\times \frac{d}{dx}\left[x\right]\right)$$

곱셈은 다르게 그룹화해도 결과가 같음.

$$\frac{d}{dx}\left[osx\right]=\frac{d}{dx}[o\times \left(sx\right)]$$

도함수의 곱의 규칙을 적용합니다.

$$\frac{d}{dx}[o\times \left(sx\right)]=\frac{d}{dx}\left[o\right]\times \left(sx\right)+o\times \frac{d}{dx}\left[sx\right]$$

곱셈에 대한 도함수를 확장합니다.

$$\frac{d}{dx}\left[e^c\right]\times \left(osx\right)+e^c\times \frac{d}{dx}\left[osx\right]=\frac{d}{dx}\left[e^c\right]\times \left(osx\right)+e^c\times \left(\frac{d}{dx}\left[o\right]\times sx+o\times \frac{d}{dx}\left[s\right]\times x+os\times \frac{d}{dx}\left[x\right]\right)$$

도함수의 곱의 규칙을 적용합니다.

$$\frac{d}{dx}\left[sx\right]=\frac{d}{dx}\left[s\right]\times x+s\times \frac{d}{dx}\left[x\right]$$

곱셈은 다르게 그룹화해도 결과가 같음.

$$\frac{d}{dx}\left[o\right]\times \left(sx\right)+o\left(\frac{d}{dx}\left[s\right]\times x+s\times \frac{d}{dx}\left[x\right]\right)=\frac{d}{dx}\left[o\right]\times sx+o\left(\frac{d}{dx}\left[s\right]\times x+s\times \frac{d}{dx}\left[x\right]\right)$$

두 수의 합이나 차에 숫자를 곱할 때는 각 숫자를 개별적으로 곱한 다음 결과를 더하거나 뺄 수 있습니다.

$$\frac{d}{dx}\left[o\right]\times sx+o\left(\frac{d}{dx}\left[s\right]\times x+s\times \frac{d}{dx}\left[x\right]\right)=\frac{d}{dx}\left[o\right]\times sx+\left(o\times \left(\frac{d}{dx}\left[s\right]\times x\right)+o\times \left(s\times \frac{d}{dx}\left[x\right]\right)\right)$$

곱셈은 다르게 그룹화해도 결과가 같음.

$$\frac{d}{dx}\left[o\right]\times sx+\left(o\times \left(\frac{d}{dx}\left[s\right]\times x\right)+o\times \left(s\times \frac{d}{dx}\left[x\right]\right)\right)=\frac{d}{dx}\left[o\right]\times sx+\left(o\times \frac{d}{dx}\left[s\right]\times x+o\times \left(s\times \frac{d}{dx}\left[x\right]\right)\right)$$

곱셈은 다르게 그룹화해도 결과가 같음.

$$\frac{d}{dx}\left[o\right]\times sx+\left(o\times \frac{d}{dx}\left[s\right]\times x+o\times \left(s\times \frac{d}{dx}\left[x\right]\right)\right)=\frac{d}{dx}\left[o\right]\times sx+\left(o\times \frac{d}{dx}\left[s\right]\times x+os\times \frac{d}{dx}\left[x\right]\right)$$

덧셈은 다르게 그룹화해도 결과가 같음.

$$\frac{d}{dx}\left[o\right]\times sx+\left(o\times \frac{d}{dx}\left[s\right]\times x+os\times \frac{d}{dx}\left[x\right]\right)=\frac{d}{dx}\left[o\right]\times sx+o\times \frac{d}{dx}\left[s\right]\times x+os\times \frac{d}{dx}\left[x\right]$$

곱셈은 다르게 그룹화해도 결과가 같음.

$$\frac{d}{dx}\left[e^c\right]\times \left(osx\right)+e^c\times \left(\frac{d}{dx}\left[o\right]\times sx+o\times \frac{d}{dx}\left[s\right]\times x+os\times \frac{d}{dx}\left[x\right]\right)=\frac{d}{dx}\left[e^c\right]\times osx+e^c\times \left(\frac{d}{dx}\left[o\right]\times sx+o\times \frac{d}{dx}\left[s\right]\times x+os\times \frac{d}{dx}\left[x\right]\right)$$

두 수의 합이나 차에 숫자를 곱할 때는 각 숫자를 개별적으로 곱한 다음 결과를 더하거나 뺄 수 있습니다.

$$\frac{d}{dx}\left[e^c\right]\times osx+e^c\times \left(\frac{d}{dx}\left[o\right]\times sx+o\times \frac{d}{dx}\left[s\right]\times x+os\times \frac{d}{dx}\left[x\right]\right)=\frac{d}{dx}\left[e^c\right]\times osx+\left(e^c\times \left(\frac{d}{dx}\left[o\right]\times sx\right)+e^c\times \left(o\times \frac{d}{dx}\left[s\right]\times x\right)+e^c\times \left(os\times \frac{d}{dx}\left[x\right]\right)\right)$$

곱셈은 다르게 그룹화해도 결과가 같음.

$$\frac{d}{dx}\left[e^c\right]\times osx+\left(e^c\times \left(\frac{d}{dx}\left[o\right]\times sx\right)+e^c\times \left(o\times \frac{d}{dx}\left[s\right]\times x\right)+e^c\times \left(os\times \frac{d}{dx}\left[x\right]\right)\right)=\frac{d}{dx}\left[e^c\right]\times osx+\left(e^c\times \frac{d}{dx}\left[o\right]\times sx+e^c\times \left(o\times \frac{d}{dx}\left[s\right]\times x\right)+e^c\times \left(os\times \frac{d}{dx}\left[x\right]\right)\right)$$

곱셈은 다르게 그룹화해도 결과가 같음.

$$\frac{d}{dx}\left[e^c\right]\times osx+\left(e^c\times \frac{d}{dx}\left[o\right]\times sx+e^c\times \left(o\times \frac{d}{dx}\left[s\right]\times x\right)+e^c\times \left(os\times \frac{d}{dx}\left[x\right]\right)\right)=\frac{d}{dx}\left[e^c\right]\times osx+\left(e^c\times \frac{d}{dx}\left[o\right]\times sx+e^c\times o\times \frac{d}{dx}\left[s\right]\times x+e^c\times \left(os\times \frac{d}{dx}\left[x\right]\right)\right)$$

곱셈은 다르게 그룹화해도 결과가 같음.

$$\frac{d}{dx}\left[e^c\right]\times osx+\left(e^c\times \frac{d}{dx}\left[o\right]\times sx+e^c\times o\times \frac{d}{dx}\left[s\right]\times x+e^c\times \left(os\times \frac{d}{dx}\left[x\right]\right)\right)=\frac{d}{dx}\left[e^c\right]\times osx+\left(e^c\times \frac{d}{dx}\left[o\right]\times sx+e^c\times o\times \frac{d}{dx}\left[s\right]\times x+e^c\times os\times \frac{d}{dx}\left[x\right]\right)$$

덧셈은 다르게 그룹화해도 결과가 같음.

$$\frac{d}{dx}\left[e^c\right]\times osx+\left(e^c\times \frac{d}{dx}\left[o\right]\times sx+e^c\times o\times \frac{d}{dx}\left[s\right]\times x+e^c\times os\times \frac{d}{dx}\left[x\right]\right)=\frac{d}{dx}\left[e^c\right]\times osx+e^c\times \frac{d}{dx}\left[o\right]\times sx+e^c\times o\times \frac{d}{dx}\left[s\right]\times x+e^c\times os\times \frac{d}{dx}\left[x\right]$$