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คำตอบ - Derivative

3 x^{2} + 2 x + 1

3x^{2}+2x+1

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1. ขั้นตอน 160: จัดรูปอนุพันธ์

9 additional steps

Expanding the derivative for addition.

$\frac{d}{d x} [x^{3} + x^{2} + x + 1] = \frac{d}{d x} [x^{3}] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Expanding the derivative for addition.

$\frac{d}{d x} [x^{3} + x^{2} + x + 1] = \frac{d}{d x} [x^{3}] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1]$

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

$\frac{d}{d x} [x^{3} + x^{2} + x + 1] = \frac{d}{d x} [x^{3} + (x^{2} + x + 1)]$

Applying the sum rule of derivatives.

$\frac{d}{d x} [x^{3} + (x^{2} + x + 1)] = \frac{d}{d x} [x^{3}] + \frac{d}{d x} [x^{2} + x + 1]$

Expanding the derivative for addition.

$\frac{d}{d x} [x^{3} + x^{2} + x + 1] = \frac{d}{d x} [x^{3}] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Expanding the derivative for addition.

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [x^{2} + x + 1] = \frac{d}{d x} [x^{3}] + (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1])$

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

$\frac{d}{d x} [x^{2} + x + 1] = \frac{d}{d x} [x^{2} + (x + 1)]$

Applying the sum rule of derivatives.

$\frac{d}{d x} [x^{2} + (x + 1)] = \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x + 1]$

Applying the sum rule of derivatives.

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [x + 1] = \frac{d}{d x} [x^{2}] + (\frac{d}{d x} [x] + \frac{d}{d x} [1])$

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

$\frac{d}{d x} [x^{2}] + (\frac{d}{d x} [x] + \frac{d}{d x} [1]) = \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1]$

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

$\frac{d}{d x} [x^{3}] + (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1]) = \frac{d}{d x} [x^{3}] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1]$

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

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1] = 3 x^{3 - 1} + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Subtracting one from a number.

$3 x^{3 - 1} + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1] = 3 x^{2} + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1]$

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

$3 x^{2} + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [1] = 3 x^{2} + 2 x^{2 - 1} + \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Subtracting one from a number.

$3 x^{2} + 2 x^{2 - 1} + \frac{d}{d x} [x] + \frac{d}{d x} [1] = 3 x^{2} + 2 x^{1} + \frac{d}{d x} [x] + \frac{d}{d x} [1]$

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

$3 x^{2} + 2 x^{1} + \frac{d}{d x} [x] + \frac{d}{d x} [1] = 3 x^{2} + 2 x + \frac{d}{d x} [x] + \frac{d}{d x} [1]$

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

$3 x^{2} + 2 x + \frac{d}{d x} [x] + \frac{d}{d x} [1] = 3 x^{2} + 2 x + 1 + \frac{d}{d x} [1]$

The derivative of a constant value is always zero.

$3 x^{2} + 2 x + 1 + \frac{d}{d x} [1] = 3 x^{2} + 2 x + 1 + 0$

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

$3 x^{2} + 2 x + 1 + 0 = 3 x^{2} + 2 x + 1$

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Derivative

คำตอบ - Derivative

คำอธิบายทีละขั้นตอน

1. ขั้นตอน 160: จัดรูปอนุพันธ์

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