Question

Solve ~%20" id="TexFormula1" title=" \frac{d}{dx} (2x {}^{2} - 4x + 1) \\ " alt=" \frac{d}{dx} (2x {}^{2} - 4x + 1) \\ " align="absmiddle" class="latex-formula">

thankyou ~​

Answer 1

$\huge \rm༆ Answer ༄$

Here's the solution ~

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

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

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

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

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

Answer 2

Answer:

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

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

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

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

$\sf \: = > 4x + \frac{d}{dx} ( - 4x) + \frac{d}{dx} (1)$

$\sf \: = > 4x - 4\frac{d}{dx} (x) + \frac{d}{dx} (1)$

$\sf = > 4x - 4 \times 1 + \frac{d}{dx} (1)$

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

$\sf = > 4x - 4$