Question

Differentiation of 3x²-4x+5 at x=0

Please help me someone.​

Answer 1

Answer:

the answer is -4

Step-by-step explanation:

we know that

$\frac{d}{dx} {x}^{n} = n {x}^{n - 1} \\ \frac{d}{dx} x = 1 \\ \frac{d}{dx} constant \: = 0$

5 is a constant.

now applying the above derivation rules, we get

$\frac{d}{dx} (3 {x}^{2} - 4x + 5) \\ \\ = \frac{d}{dx}(3 {x}^{2}) - \frac{d}{dx} ( 4x) + \frac{d}{dx} (5) \\ = 3 \times 2 {x}^{2 - 1} - 4 \times 1 + 0 \\ = 6x - 4 \\ \\ \\ \\ now \: at \: x = 0 \: \\ = 6 \times 0 - 4 \\ = - 4$

Answer 2

$= {3x}^{2} - 4x + 5$

$= 3 {(0)}^{2} - 4(0) + 5$

$= 3(0) - 0 + 5$

$= 0 - 0 + 5$

$= 5$