Question

Find the derivative by first principle:
3x² + 4

Answer 1

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Answer 2

Answer:

6x

Step-by-step explanation:

$f(x) = 3x^2 + 4\\f(x + h) = 3(x+h)^2 + 4 = 3(x^2 + h^2 +2xh) + 4\\ f(x+h) = 3x^2 + 3h^2 +6xh +4$

Now we need to find

$\lim_{h \to 0} \dfrac{f(x + h)- f(x)}{h}\\ \lim_{h \to 0} \dfrac {(3x^2 + 3h^2 +6xh +4) - (3x^2 + 4)}{h}\\ \lim_{h \to 0}\dfrac {3x^2 + 3h^2 +6xh +4 - 3x^2 - 4}{h}\\ \lim_{h \to 0} \dfrac {3h^2 + 6xh}h\\  \lim_{h \to 0} \dfrac {3h^2}{h} + \lim_{h \to 0} \dfrac{6xh}h\\\\\lim_{h \to 0} 3h + \lim_{h \to 0} 6x = 6x$