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Řešení - Derivative

- \frac{5 \cdot (15 x + 3)}{{(5 x + 1)}^{5}}

- \frac{5 \cdot \left(15 x + 3\right)}{\left(5 x + 1\right)^{5}}

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1. Krok 160: upravte derivaci

Computing the derivative of a power function.

$\frac{d}{d x} [{(\frac{1}{5 x + 1})}^{3}] = {(\frac{1}{5 x + 1})}^{3} \times (\frac{d}{d x} [3] \times \ln (\frac{1}{5 x + 1}) + \frac{3}{\frac{1}{5 x + 1}} \times \frac{d}{d x} [\frac{1}{5 x + 1}])$

The derivative of a constant value is always zero.

${(\frac{1}{5 x + 1})}^{3} \times (\frac{d}{d x} [3] \times \ln (\frac{1}{5 x + 1}) + \frac{3}{\frac{1}{5 x + 1}} \times \frac{d}{d x} [\frac{1}{5 x + 1}]) = {(\frac{1}{5 x + 1})}^{3} \times (0 \times \ln (\frac{1}{5 x + 1}) + \frac{3}{\frac{1}{5 x + 1}} \times \frac{d}{d x} [\frac{1}{5 x + 1}])$

Computing the derivative of a fraction.

${(\frac{1}{5 x + 1})}^{3} \times (0 \times \ln (\frac{1}{5 x + 1}) + \frac{3}{\frac{1}{5 x + 1}} \times \frac{d}{d x} [\frac{1}{5 x + 1}]) = {(\frac{1}{5 x + 1})}^{3} \times (0 \times \ln (\frac{1}{5 x + 1}) + \frac{3}{\frac{1}{5 x + 1}} \times \frac{\frac{d}{d x} [1] \times (5 x + 1) - 1 \times \frac{d}{d x} [5 x + 1]}{{(5 x + 1)}^{2}})$

Multiplying a number by zero always results in zero.

${(\frac{1}{5 x + 1})}^{3} \times (0 \times \ln (\frac{1}{5 x + 1}) + \frac{3}{\frac{1}{5 x + 1}} \times \frac{\frac{d}{d x} [1] \times (5 x + 1) - 1 \times \frac{d}{d x} [5 x + 1]}{{(5 x + 1)}^{2}}) = {(\frac{1}{5 x + 1})}^{3} \times (0 + \frac{3}{\frac{1}{5 x + 1}} \times \frac{\frac{d}{d x} [1] \times (5 x + 1) - 1 \times \frac{d}{d x} [5 x + 1]}{{(5 x + 1)}^{2}})$

The derivative of a constant value is always zero.

${(\frac{1}{5 x + 1})}^{3} \times (0 + \frac{3}{\frac{1}{5 x + 1}} \times \frac{\frac{d}{d x} [1] \times (5 x + 1) - 1 \times \frac{d}{d x} [5 x + 1]}{{(5 x + 1)}^{2}}) = {(\frac{1}{5 x + 1})}^{3} \times (0 + \frac{3}{\frac{1}{5 x + 1}} \times \frac{0 (5 x + 1) - 1 \times \frac{d}{d x} [5 x + 1]}{{(5 x + 1)}^{2}})$

Applying the sum rule of derivatives.

${(\frac{1}{5 x + 1})}^{3} \times (0 + \frac{3}{\frac{1}{5 x + 1}} \times \frac{0 (5 x + 1) - 1 \times \frac{d}{d x} [5 x + 1]}{{(5 x + 1)}^{2}}) = {(\frac{1}{5 x + 1})}^{3} \times (0 + \frac{3}{\frac{1}{5 x + 1}} \times \frac{0 (5 x + 1) - 1 (\frac{d}{d x} [5 x] + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}})$

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

${(\frac{1}{5 x + 1})}^{3} \times (0 + \frac{3}{\frac{1}{5 x + 1}} \times \frac{0 (5 x + 1) - 1 (\frac{d}{d x} [5 x] + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}}) = {(\frac{1}{5 x + 1})}^{3} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 (5 x + 1) - 1 (\frac{d}{d x} [5 x] + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}})$

Simplifying the arithmetic expressions.

${(\frac{1}{5 x + 1})}^{3} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 (5 x + 1) - 1 (\frac{d}{d x} [5 x] + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 (5 x + 1) - 1 (\frac{d}{d x} [5 x] + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}})$

Multiplying a number by zero always results in zero.

$\frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 (5 x + 1) - 1 (\frac{d}{d x} [5 x] + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 (\frac{d}{d x} [5 x] + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}})$

Applying the product rule of derivatives.

$\frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 (\frac{d}{d x} [5 x] + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 ((\frac{d}{d x} [5] \times x + 5 \times \frac{d}{d x} [x]) + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}})$

The derivative of a constant value is always zero.

$\frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 ((\frac{d}{d x} [5] \times x + 5 \times \frac{d}{d x} [x]) + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 ((0 x + 5 \times \frac{d}{d x} [x]) + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}})$

Multiplying a number by zero always results in zero.

$\frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 ((0 x + 5 \times \frac{d}{d x} [x]) + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 ((0 + 5 \times \frac{d}{d x} [x]) + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}})$

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

$\frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 ((0 + 5 \times \frac{d}{d x} [x]) + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 (5 \times \frac{d}{d x} [x] + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}})$

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

$\frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 (5 \times \frac{d}{d x} [x] + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 (5 \times 1 + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}})$

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

$\frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 (5 \times 1 + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 (5 + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}})$

The derivative of a constant value is always zero.

$\frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 (5 + \frac{d}{d x} [1])}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 (5 + 0)}{{(5 x + 1)}^{2}})$

Simplifying the arithmetic expressions.

$\frac{1}{{(5 x + 1)}^{3}} \times (\frac{3}{\frac{1}{5 x + 1}} \times \frac{0 - 1 (5 + 0)}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (15 x + 3 \times \frac{0 - 1 (5 + 0)}{{(5 x + 1)}^{2}})$

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

$\frac{1}{{(5 x + 1)}^{3}} \times (15 x + 3 \times \frac{0 - 1 (5 + 0)}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (15 x + 3 \times \frac{- 1 (5 + 0)}{{(5 x + 1)}^{2}})$

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

$\frac{1}{{(5 x + 1)}^{3}} \times (15 x + 3 \times \frac{- 1 (5 + 0)}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (15 x + 3 \times \frac{- 1 \times 5}{{(5 x + 1)}^{2}})$

Multiplying two integers together.

$\frac{1}{{(5 x + 1)}^{3}} \times (15 x + 3 \times \frac{- 1 \times 5}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (15 x + 3 \times \frac{- 5}{{(5 x + 1)}^{2}})$

Simplifying the arithmetic expressions.

$\frac{1}{{(5 x + 1)}^{3}} \times (15 x + 3 \times \frac{- 5}{{(5 x + 1)}^{2}}) = \frac{1}{{(5 x + 1)}^{3}} \times (15 x + 3 \times (- \frac{5}{{(5 x + 1)}^{2}}))$

Simplifying the arithmetic expressions.

$\frac{1}{{(5 x + 1)}^{3}} \times (15 x + 3 \times (- \frac{5}{{(5 x + 1)}^{2}})) = \frac{1}{{(5 x + 1)}^{3}} \times (- \frac{5 \cdot (15 x + 3)}{{(5 x + 1)}^{2}})$