Megoldás - Derivált

- \frac{12}{x^{2}}

- \frac{12}{x^{2}}

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1. Számítsa ki a deriváltat

9 additional steps

Expanding the derivative for multiplication.

$\frac{d}{d x} [3 x \times \frac{4}{x^{2}}] = \frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + 3 \times \frac{d}{d x} [x] \times \frac{4}{x^{2}} + 3 x \times \frac{d}{d x} [\frac{4}{x^{2}}]$

Expanding the derivative for multiplication.

$\frac{d}{d x} [3 x \times \frac{4}{x^{2}}] = \frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + 3 \times \frac{d}{d x} [x] \times \frac{4}{x^{2}} + 3 x \times \frac{d}{d x} [\frac{4}{x^{2}}]$

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

$\frac{d}{d x} [3 x \times \frac{4}{x^{2}}] = \frac{d}{d x} [3 \times (x \times \frac{4}{x^{2}})]$

Applying the product rule of derivatives.

$\frac{d}{d x} [3 \times (x \times \frac{4}{x^{2}})] = \frac{d}{d x} [3] \times (x \times \frac{4}{x^{2}}) + 3 \times \frac{d}{d x} [x \times \frac{4}{x^{2}}]$

Expanding the derivative for multiplication.

$\frac{d}{d x} [3 x \times \frac{4}{x^{2}}] = \frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + 3 \times \frac{d}{d x} [x] \times \frac{4}{x^{2}} + 3 x \times \frac{d}{d x} [\frac{4}{x^{2}}]$

Applying the product rule of derivatives.

$\frac{d}{d x} [x \times \frac{4}{x^{2}}] = \frac{d}{d x} [x] \times \frac{4}{x^{2}} + x \times \frac{d}{d x} [\frac{4}{x^{2}}]$

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

$\frac{d}{d x} [3] \times (x \times \frac{4}{x^{2}}) + 3 (\frac{d}{d x} [x] \times \frac{4}{x^{2}} + x \times \frac{d}{d x} [\frac{4}{x^{2}}]) = \frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + 3 (\frac{d}{d x} [x] \times \frac{4}{x^{2}} + x \times \frac{d}{d x} [\frac{4}{x^{2}}])$

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.

$\frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + 3 (\frac{d}{d x} [x] \times \frac{4}{x^{2}} + x \times \frac{d}{d x} [\frac{4}{x^{2}}]) = \frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + (3 \times (\frac{d}{d x} [x] \times \frac{4}{x^{2}}) + 3 \times (x \times \frac{d}{d x} [\frac{4}{x^{2}}]))$

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

$\frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + (3 \times (\frac{d}{d x} [x] \times \frac{4}{x^{2}}) + 3 \times (x \times \frac{d}{d x} [\frac{4}{x^{2}}])) = \frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + (3 \times \frac{d}{d x} [x] \times \frac{4}{x^{2}} + 3 \times (x \times \frac{d}{d x} [\frac{4}{x^{2}}]))$

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

$\frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + (3 \times \frac{d}{d x} [x] \times \frac{4}{x^{2}} + 3 \times (x \times \frac{d}{d x} [\frac{4}{x^{2}}])) = \frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + (3 \times \frac{d}{d x} [x] \times \frac{4}{x^{2}} + 3 x \times \frac{d}{d x} [\frac{4}{x^{2}}])$

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

$\frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + (3 \times \frac{d}{d x} [x] \times \frac{4}{x^{2}} + 3 x \times \frac{d}{d x} [\frac{4}{x^{2}}]) = \frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + 3 \times \frac{d}{d x} [x] \times \frac{4}{x^{2}} + 3 x \times \frac{d}{d x} [\frac{4}{x^{2}}]$

The derivative of a constant value is always zero.

$\frac{d}{d x} [3] \times x \times \frac{4}{x^{2}} + 3 \times \frac{d}{d x} [x] \times \frac{4}{x^{2}} + 3 x \times \frac{d}{d x} [\frac{4}{x^{2}}] = 0 x \times \frac{4}{x^{2}} + 3 \times \frac{d}{d x} [x] \times \frac{4}{x^{2}} + 3 x \times \frac{d}{d x} [\frac{4}{x^{2}}]$

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

$0 x \times \frac{4}{x^{2}} + 3 \times \frac{d}{d x} [x] \times \frac{4}{x^{2}} + 3 x \times \frac{d}{d x} [\frac{4}{x^{2}}] = 0 x \times \frac{4}{x^{2}} + 3 \times 1 \times \frac{4}{x^{2}} + 3 x \times \frac{d}{d x} [\frac{4}{x^{2}}]$

Computing the derivative of a fraction.

$0 x \times \frac{4}{x^{2}} + 3 \times 1 \times \frac{4}{x^{2}} + 3 x \times \frac{d}{d x} [\frac{4}{x^{2}}] = 0 x \times \frac{4}{x^{2}} + 3 \times 1 \times \frac{4}{x^{2}} + 3 x \times \frac{\frac{d}{d x} [4] \times x^{2} - 4 \times \frac{d}{d x} [x^{2}]}{{(x^{2})}^{2}}$

The derivative of a constant value is always zero.

$0 x \times (\frac{4}{x^{2}}) + 3 \times 1 \times (\frac{4}{x^{2}}) + 3 x \times \frac{\frac{d}{d x} [4] \times x^{2} - 4 \times \frac{d}{d x} [x^{2}]}{{(x^{2})}^{2}} = 0 x \times (\frac{4}{x^{2}}) + 3 \times 1 \times (\frac{4}{x^{2}}) + 3 x \times \frac{0 x^{2} - 4 \times \frac{d}{d x} [x^{2}]}{{(x^{2})}^{2}}$

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

$0 x \times (\frac{4}{x^{2}}) + 3 \times 1 \times (\frac{4}{x^{2}}) + 3 x \times \frac{0 x^{2} - 4 \times \frac{d}{d x} [x^{2}]}{{(x^{2})}^{2}} = 0 x \times (\frac{4}{x^{2}}) + 3 \times 1 \times (\frac{4}{x^{2}}) + 3 x \times \frac{0 x^{2} - 4 \times (2 x^{2 - 1})}{{(x^{2})}^{2}}$

Subtracting one from a number.

$0 x \times (\frac{4}{x^{2}}) + 3 \times 1 \times (\frac{4}{x^{2}}) + 3 x \times \frac{0 x^{2} - 4 \times (2 x^{2 - 1})}{{(x^{2})}^{2}} = 0 x \times (\frac{4}{x^{2}}) + 3 \times 1 \times (\frac{4}{x^{2}}) + 3 x \times \frac{0 x^{2} - 4 \times (2 x^{1})}{{(x^{2})}^{2}}$

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

$0 x \times (\frac{4}{x^{2}}) + 3 \times 1 \times (\frac{4}{x^{2}}) + 3 x \times \frac{0 x^{2} - 4 \times (2 x^{1})}{{(x^{2})}^{2}} = 0 x \times (\frac{4}{x^{2}}) + 3 \times 1 \times (\frac{4}{x^{2}}) + 3 x \times \frac{0 x^{2} - 4 \times (2 x)}{{(x^{2})}^{2}}$