คำตอบ - Derivative

Expanding the derivative for multiplication.

$$\frac{d}{dx}[s\times e\times n\times \left(10x^2\right)]=\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+s\times \frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+s\times e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+s\times e\times n\times \frac{d}{dx}\left[10x^2\right]$$

Expanding the derivative for multiplication.

$$\frac{d}{dx}[s\times e\times n\times \left(10x^2\right)]=\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+s\times \frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+s\times e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+s\times e\times n\times \frac{d}{dx}\left[10x^2\right]$$

Multiplication can be grouped differently, but the result remains the same.

$$\frac{d}{dx}[s\times e\times n\times \left(10x^2\right)]=\frac{d}{dx}[s\times \left(e\times n\times \left(10x^2\right)\right)]$$

Applying the product rule of derivatives.

$$\frac{d}{dx}[s\times \left(e\times n\times \left(10x^2\right)\right)]=\frac{d}{dx}\left[s\right]\times \left(e\times n\times \left(10x^2\right)\right)+s\times \frac{d}{dx}[e\times n\times \left(10x^2\right)]$$

Expanding the derivative for multiplication.

$$\frac{d}{dx}[s\times e\times n\times \left(10x^2\right)]=\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+s\times \frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+s\times e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+s\times e\times n\times \frac{d}{dx}\left[10x^2\right]$$

Expanding the derivative for multiplication.

$$\frac{d}{dx}\left[s\right]\times \left(e\times n\times \left(10x^2\right)\right)+s\times \frac{d}{dx}[e\times n\times \left(10x^2\right)]=\frac{d}{dx}\left[s\right]\times \left(e\times n\times \left(10x^2\right)\right)+s\left(\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+e\times n\times \frac{d}{dx}\left[10x^2\right]\right)$$

Multiplication can be grouped differently, but the result remains the same.

$$\frac{d}{dx}[e\times n\times \left(10x^2\right)]=\frac{d}{dx}[e\times \left(n\times \left(10x^2\right)\right)]$$

Applying the product rule of derivatives.

$$\frac{d}{dx}[e\times \left(n\times \left(10x^2\right)\right)]=\frac{d}{dx}\left[e\right]\times \left(n\times \left(10x^2\right)\right)+e\times \frac{d}{dx}[n\times \left(10x^2\right)]$$

Expanding the derivative for multiplication.

$$\frac{d}{dx}\left[s\right]\times \left(e\times n\times \left(10x^2\right)\right)+s\times \frac{d}{dx}[e\times n\times \left(10x^2\right)]=\frac{d}{dx}\left[s\right]\times \left(e\times n\times \left(10x^2\right)\right)+s\left(\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+e\times n\times \frac{d}{dx}\left[10x^2\right]\right)$$

Applying the product rule of derivatives.

$$\frac{d}{dx}[n\times \left(10x^2\right)]=\frac{d}{dx}\left[n\right]\times \left(10x^2\right)+n\times \frac{d}{dx}\left[10x^2\right]$$

Multiplication can be grouped differently, but the result remains the same.

$$\frac{d}{dx}\left[e\right]\times \left(n\times \left(10x^2\right)\right)+e\times \left(\frac{d}{dx}\left[n\right]\times \left(10x^2\right)+n\times \frac{d}{dx}\left[10x^2\right]\right)=\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+e\times \left(\frac{d}{dx}\left[n\right]\times \left(10x^2\right)+n\times \frac{d}{dx}\left[10x^2\right]\right)$$

Multiplying a number by a sum or difference of two numbers can be done by multiplying each number individually and then adding or subtracting the results.

$$\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+e\times \left(\frac{d}{dx}\left[n\right]\times \left(10x^2\right)+n\times \frac{d}{dx}\left[10x^2\right]\right)=\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+\left(e\times \left(\frac{d}{dx}\left[n\right]\times \left(10x^2\right)\right)+e\times \left(n\times \frac{d}{dx}\left[10x^2\right]\right)\right)$$

Multiplication can be grouped differently, but the result remains the same.

$$\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+\left(e\times \left(\frac{d}{dx}\left[n\right]\times \left(10x^2\right)\right)+e\times \left(n\times \frac{d}{dx}\left[10x^2\right]\right)\right)=\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+\left(e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+e\times \left(n\times \frac{d}{dx}\left[10x^2\right]\right)\right)$$

Multiplication can be grouped differently, but the result remains the same.

$$\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+\left(e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+e\times \left(n\times \frac{d}{dx}\left[10x^2\right]\right)\right)=\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+\left(e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+e\times n\times \frac{d}{dx}\left[10x^2\right]\right)$$

Addition can be grouped differently, but the result remains the same.

$$\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+\left(e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+e\times n\times \frac{d}{dx}\left[10x^2\right]\right)=\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+e\times n\times \frac{d}{dx}\left[10x^2\right]$$

Multiplication can be grouped differently, but the result remains the same.

$$\frac{d}{dx}\left[s\right]\times \left(e\times n\times \left(10x^2\right)\right)+s\left(\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+e\times n\times \frac{d}{dx}\left[10x^2\right]\right)=\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+s\left(\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+e\times n\times \frac{d}{dx}\left[10x^2\right]\right)$$

Multiplying a number by a sum or difference of two numbers can be done by multiplying each number individually and then adding or subtracting the results.

$$\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+s\left(\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+e\times n\times \frac{d}{dx}\left[10x^2\right]\right)=\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+\left(s\times \left(\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)\right)+s\times \left(e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)\right)+s\times \left(e\times n\times \frac{d}{dx}\left[10x^2\right]\right)\right)$$

Multiplication can be grouped differently, but the result remains the same.

$$\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+\left(s\times \left(\frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)\right)+s\times \left(e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)\right)+s\times \left(e\times n\times \frac{d}{dx}\left[10x^2\right]\right)\right)=\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+\left(s\times \frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+s\times \left(e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)\right)+s\times \left(e\times n\times \frac{d}{dx}\left[10x^2\right]\right)\right)$$

Multiplication can be grouped differently, but the result remains the same.

$$\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+\left(s\times \frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+s\times \left(e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)\right)+s\times \left(e\times n\times \frac{d}{dx}\left[10x^2\right]\right)\right)=\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+\left(s\times \frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+s\times e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+s\times \left(e\times n\times \frac{d}{dx}\left[10x^2\right]\right)\right)$$

Multiplication can be grouped differently, but the result remains the same.

$$\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+\left(s\times \frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+s\times e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+s\times \left(e\times n\times \frac{d}{dx}\left[10x^2\right]\right)\right)=\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+\left(s\times \frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+s\times e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+s\times e\times n\times \frac{d}{dx}\left[10x^2\right]\right)$$

Addition can be grouped differently, but the result remains the same.

$$\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+\left(s\times \frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+s\times e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+s\times e\times n\times \frac{d}{dx}\left[10x^2\right]\right)=\frac{d}{dx}\left[s\right]\times e\times n\times \left(10x^2\right)+s\times \frac{d}{dx}\left[e\right]\times n\times \left(10x^2\right)+s\times e\times \frac{d}{dx}\left[n\right]\times \left(10x^2\right)+s\times e\times n\times \frac{d}{dx}\left[10x^2\right]$$