Solve ~ (d)/(dx) (2x {}^(2) - 4x + 1) \\ thankyou ~

Question 1

Solve ~

$(d)/(dx) (2x {}^(2) - 4x + 1) \\$

thankyou ~

Question 2

$\huge \rm༆ Answer ༄$

Here's the solution ~

${ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: (d)/(dx) (2 {x}^(2) - 4x + 1)$

${ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: (d)/(dx) (2 {x}^(2)) - (d)/(dx) ( 4x )+ (d)/(dx) (1)$

${ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: (2 * 2x {}^(2 - 1) { }^{}) - ( 1 * 4x ^ {1 - 1})+ (0)$

${ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: (2 * 2x {}^{} { }^{}) - ( 1 * 4 ^ {})+ 0$

${ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:4x - 4$

Question 3

Answer:

$\sf = > (d)/(dx) ( {2x}^(2) - 4x + 1)$

$\sf = > (d)/(dx) ( {2x}^(2) ) +(d)/(dx) ( - 4x) + (d)/(dx) (1)$

$\sf = > 2(d)/(dx) ( {x}^(2) ) + (d)/(dx) ( - 4x) + (d)/(dx) (1)$

$\sf= > 2(2x) + (d)/(dx) ( - 4x) + (d)/(dx) (1)$

$\sf \: = > 4x + (d)/(dx) ( - 4x) + (d)/(dx) (1)$

$\sf \: = > 4x - 4(d)/(dx) (x) + (d)/(dx) (1)$

$\sf = > 4x - 4 * 1 + (d)/(dx) (1)$

$\sf \: = > 4x - 4 + 0$

$\sf = > 4x - 4$

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