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Can someone help me with basic derivatives: 3x^2 + 5x + 5?

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6x + 5
3(2X) + 5

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Other answers:

The rules are: derivative of x^n = n * x^(n-1) derivative of sum is the sum of the derivatives. derivative of a constant times a function equals the constant times the derivative of the function.
The approach is pretty straightforward -- \(x^2\) on its own derives to 2x. Since it has a 3 in front of it, we get 3(2x), as above. x derives to \(x^0\), which is the same as 1. Since it has a 5 in front of it, we get 5, as above. 5 disappears, because it is a constant and the derivative of a constant is 0.
So d/dx (3x^2 +5x + 5) = d/dx (3x^2) + d/dx (5x) + d/dx (5) = 3* d/dx (x^2) + 5 * d/dx (x) + 0 = 3 * 2 x + 5 * 1 = 6x + 5
\[\lim_{h\rightarrow 0}\ \frac{f(x+h)-f(x)}{h}\] \[(3(x+h)^2 +5(x+h) +5)-(3x^2 +5x +5)//h\] \[(3(x^2+2xh+h^2) +5x+5h +5 -3x^2 -5x -5)//h\] \[3x^2+6xh+3h^2 +5x+5h +5 -3x^2 -5x -5)//h\] \[\frac{6xh+3h^2+5h}{h}\] \[\lim_{h\rightarrow0}\ 6x+3h+5\ =\ 6x+5\]
the long version lets you appreciate the shortcut rules ;)

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