solve ~
[tex] \frac{d}{dx} (2x {}^{2} - 4x + 1) \\ [/tex]

thankyou ~

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DᴀʀᴋPᴀʀᴀᴅᴏx DᴀʀᴋPᴀʀᴀᴅᴏx · Answer 1 · 2022-01-22T09:04:42+01:00

[tex]\huge \rm༆ Answer ༄[/tex]

Here's the solution ~

[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: \dfrac{d}{dx} (2 {x}^{2} - 4x + 1)[/tex]

[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: \dfrac{d}{dx} (2 {x}^{2}) - \dfrac{d}{dx} ( 4x )+ \dfrac{d}{dx} (1)[/tex]

[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: (2 \times 2x {}^{2 - 1} { }^{}) - ( 1 \times 4x ^ {1 - 1})+ (0)[/tex]

[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: (2 \times 2x {}^{} { }^{}) - ( 1 \times 4 ^ {})+ 0[/tex]

[tex]{ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:4x - 4[/tex]

Аноним Аноним · Answer 2 · 2022-01-23T04:54:30+01:00

Answer:

[tex] \sf = > \frac{d}{dx} ( {2x}^{2} - 4x + 1)[/tex]

[tex] \sf = > \frac{d}{dx} ( {2x}^{2} ) +\frac{d}{dx} ( - 4x) + \frac{d}{dx} (1)[/tex]

[tex] \sf = > 2\frac{d}{dx} ( {x}^{2} ) + \frac{d}{dx} ( - 4x) + \frac{d}{dx} (1)[/tex]

[tex] \sf= > 2(2x) + \frac{d}{dx} ( - 4x) + \frac{d}{dx} (1)[/tex]

[tex] \sf \: = > 4x + \frac{d}{dx} ( - 4x) + \frac{d}{dx} (1)[/tex]

[tex] \sf \: = > 4x - 4\frac{d}{dx} (x) + \frac{d}{dx} (1)[/tex]

[tex] \sf = > 4x - 4 \times 1 + \frac{d}{dx} (1)[/tex]

[tex] \sf \: = > 4x - 4 + 0[/tex]

[tex] \sf = > 4x - 4[/tex]