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192 x^{2} {(2 x^{2} + 5)}^{2} + 16 {(2 x^{2} + 5)}^{3}

192 x^{2} \left(2 x^{2} + 5\right)^{2}+16 \left(2 x^{2} + 5\right)^{3}

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Maelezo kwa hatua

1. Suluhisha johofu

9 ziada steps

Kupanua kifungu kwa ajili ya bidhaa.

$\frac{d}{d x} [4 {(2 x^{2} + 5)}^{3} \times (4 x)] = \frac{d}{d x} [4] \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times \frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x]$

Kupanua kifungu kwa ajili ya bidhaa.

Uzidishaji unaweza kikundi tofauti, lakini matokeo yanabaki yale yale.

$\frac{d}{d x} [4 {(2 x^{2} + 5)}^{3} \times (4 x)] = \frac{d}{d x} [4 \times ({(2 x^{2} + 5)}^{3} \times (4 x))]$

Kutumia sheria ya bidhaa ya vifungu.

$\frac{d}{d x} [4 \times ({(2 x^{2} + 5)}^{3} \times (4 x))] = \frac{d}{d x} [4] \times ({(2 x^{2} + 5)}^{3} \times (4 x)) + 4 \times \frac{d}{d x} [{(2 x^{2} + 5)}^{3} \times (4 x)]$

Kupanua kifungu kwa ajili ya bidhaa.

Kutumia sheria ya bidhaa ya vifungu.

$\frac{d}{d x} [{(2 x^{2} + 5)}^{3} \times (4 x)] = \frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x]$

Uzidishaji unaweza kikundi tofauti, lakini matokeo yanabaki yale yale.

$\frac{d}{d x} [4] \times ({(2 x^{2} + 5)}^{3} \times (4 x)) + 4 (\frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x]) = \frac{d}{d x} [4] \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 (\frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x])$

Kuzidisha namba na jumla au tofauti ya namba mbili inaweza kufanyika kwa kuzidisha kila namba pekee na kisha kuongeza au kupunguza matokeo.

$\frac{d}{d x} [4] \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 (\frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x]) = \frac{d}{d x} [4] \times {(2 x^{2} + 5)}^{3} \times (4 x) + (4 \times (\frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x)) + 4 \times ({(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x]))$

Uzidishaji unaweza kikundi tofauti, lakini matokeo yanabaki yale yale.

$\frac{d}{d x} [4] \times {(2 x^{2} + 5)}^{3} \times (4 x) + (4 \times (\frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x)) + 4 \times ({(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x])) = \frac{d}{d x} [4] \times {(2 x^{2} + 5)}^{3} \times (4 x) + (4 \times \frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x]))$

Uzidishaji unaweza kikundi tofauti, lakini matokeo yanabaki yale yale.

$\frac{d}{d x} [4] \times {(2 x^{2} + 5)}^{3} \times (4 x) + (4 \times \frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x])) = \frac{d}{d x} [4] \times {(2 x^{2} + 5)}^{3} \times (4 x) + (4 \times \frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x])$

Jumla zinaweza kikundi tofauti, lakini matokeo yanabaki yale yale.

$\frac{d}{d x} [4] \times {(2 x^{2} + 5)}^{3} \times (4 x) + (4 \times \frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x]) = \frac{d}{d x} [4] \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times \frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x]$

Mchanganyiko wa mabadiliko ya thamani ya kudumu ni sawa na sifuri.

$\frac{d}{d x} [4] \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times \frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x] = 0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times \frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x]$

Kuhesabu kifungu cha kazi ya uwezo.

$0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times \frac{d}{d x} [{(2 x^{2} + 5)}^{3}] \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x] = 0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x]$

Kutumia sheria ya bidhaa ya vifungu.

$0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times \frac{d}{d x} [4 x] = 0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} (\frac{d}{d x} [4] \times x + 4 \times \frac{d}{d x} [x])$

Mchanganyiko wa mabadiliko ya thamani ya kudumu ni sawa na sifuri.

$0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} (\frac{d}{d x} [4] \times x + 4 \times \frac{d}{d x} [x]) = 0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} (0 x + 4 \times \frac{d}{d x} [x])$

Kuzidisha idadi kwa sifuri daima hutoa sifuri.

$0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} (0 x + 4 \times \frac{d}{d x} [x]) = 0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} (0 + 4 \times \frac{d}{d x} [x])$

Kuongeza sifuri kwa idadi, ambayo haibadilishi thamani yake.

$0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} (0 + 4 \times \frac{d}{d x} [x]) = 0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times (4 \times \frac{d}{d x} [x])$

Mchanganyiko wa mabadiliko ya kigezo kulingana na yenyewe ni sawa na moja kila wakati.

$0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times (4 \times \frac{d}{d x} [x]) = 0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times (4 \times 1)$

Kuzidisha idadi na moja, ambayo haibadilishi thamani yake.

$0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times (4 \times 1) = 0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Mchanganyiko wa mabadiliko ya thamani ya kudumu ni sawa na sifuri.

$0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (\frac{d}{d x} [3] \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Kutumia sheria ya jumla ya vifungu.

$0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times \frac{d}{d x} [2 x^{2} + 5])) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times (\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Kuzidisha idadi kwa sifuri daima hutoa sifuri.

$0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 \times \ln (2 x^{2} + 5) + \frac{3}{2 x^{2} + 5} \times (\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times (\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Kutumia sheria ya bidhaa ya vifungu.

$0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times (\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times ((\frac{d}{d x} [2] \times x^{2} + 2 \times \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Mchanganyiko wa mabadiliko ya thamani ya kudumu ni sawa na sifuri.

$0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times ((\frac{d}{d x} [2] \times x^{2} + 2 \times \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times ((0 x^{2} + 2 \times \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Kuzidisha idadi kwa sifuri daima hutoa sifuri.

$0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times ((0 x^{2} + 2 \times \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times ((0 + 2 \times \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Kuongeza sifuri kwa idadi, ambayo haibadilishi thamani yake.

$0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times ((0 + 2 \times \frac{d}{d x} [x^{2}]) + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times (2 \times \frac{d}{d x} [x^{2}] + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Kuhesabu kifungu cha x iliyoinuka kwa nguvu ya n.

$0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times (2 \times \frac{d}{d x} [x^{2}] + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times (2 \times (2 x^{2 - 1}) + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Kupunguza moja kutoka kwenye namba.

$0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times (2 \times (2 x^{2 - 1}) + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times (2 \times (2 x^{1}) + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Uzidishaji unaweza kikundi tofauti, lakini matokeo yanabaki yale yale.

$0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times (2 \times (2 x^{1}) + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times ((2 \times 2) \times x^{1} + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Kuzidisha vigezo mbili pamoja.

$0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times ((2 \times 2) \times x^{1} + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times (4 x^{1} + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Mchanganyiko wa mabadiliko ya thamani ya kudumu ni sawa na sifuri.

$0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times (4 x^{1} + \frac{d}{d x} [5]))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4 = 0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} (0 + \frac{3}{2 x^{2} + 5} \times (4 x^{1} + 0))) \times (4 x) + 4 {(2 x^{2} + 5)}^{3} \times 4$

Kuongeza sifuri kwa idadi, ambayo haibadilishi thamani yake.

$0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} \times (0 + \frac{3}{2 x^{2} + 5} \times (4 x^{1} + 0))) \times (4 x) + 4 \times {(2 x^{2} + 5)}^{3} \times 4 = 0 \times {(2 x^{2} + 5)}^{3} \times (4 x) + 4 \times ({(2 x^{2} + 5)}^{3} \times (\frac{3}{2 x^{2} + 5} \times (4 x^{1} + 0))) \times (4 x) + 4 \times {(2 x^{2} + 5)}^{3} \times 4$