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Dimas [21]
4 years ago
12

Find the derivative of f(x) = 6x + 2 at x = 1. 1 2 3 6

Mathematics
2 answers:
noname [10]4 years ago
7 0

f(x)=6x+2,\ x_0=1\\\\f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(x_0)}{h}\\\\f'(1)=\lim\limits_{h\to0}\dfrac{f(1+h)-f(1)}{h}=\lim\limits_{h\to0}\dfrac{6(1+h)+2-[6(1)+2]}{h}\\\\=\lim\limits_{h\to0}\dfrac{(6)(1)+(6)(h)-(6+2)}{h}=\lim\limits_{h\to0}\dfrac{6+6h-8}{h}\\\\=\lim\limits_{h\to0}\dfrac{6h-2}{h}=\lim\limits_{h\to0}\left(\dfrac{6h}{h}-\dfrac{2}{h}\right)=\lim\limits_{h\to0}\left(6-\dfrac{2}{h}\right)=6\\\\Answer:\ \boxed{f'(1)=6}

marysya [2.9K]4 years ago
4 0
Ich would das that the derivative of
f(x) = 6x+2
is 6
because of derivative 6x it is 6 and the derivative of 2 is 0
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4 years ago
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alexandr1967 [171]
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6 0
3 years ago
Long division<br> 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2
inna [77]
(6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2) / (3x+1)

We divide first number from first parenthesis with first number from second parenthesis. Then the resulting number we multiply by all numbers in second parenthesis and substract from first parenthesis.

6 x^{6}/3x = 2 x^{5} \\  \\ 6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2 - 6 x^{6} -2 x^{5} = 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2\\

We repeat previous steps until we run out of numbers:
3x^{5}/3x=x^{4} \\ \\ 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2-3x^{5}-x^{4}= \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2 \\ \\ \\ x^{4}/3x= \frac{1}{3} x^{3} \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2-x^{4}- \frac{1}{3} x^{3}= \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2 \\ \\ \\ - \frac{28}{3} x^{3}/3x= - \frac{28}{9} x^{2} \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2+ \frac{28}{3} x^{3}+ \frac{28}{9} x^{2} = \\ \\ \frac{91}{9} x^{2}-10x+2
\frac{91}{9} x^{2}/3x=\frac{91}{27} x \\ \\ \frac{91}{9} x^{2}-10x+2-\frac{91}{9} x^{2}-\frac{91}{27} x= \\ \\ -\frac{361}{27} x+2 \\ \\ \\ -\frac{361}{27} x/3x=-\frac{361}{81} \\ \\ -\frac{361}{27} x+2+\frac{361}{27}x+\frac{361}{27}= \\ \\ \frac{415}{27}

We are left with a number that has no x inside. This is remainder.
The final solution is sum of all these solutions and remainder:
(2 x^{5}+x^{4}+\frac{1}{3} x^{3} - \frac{28}{9} x^{2} +\frac{91}{27} x)+(-\frac{361}{81} )
6 0
3 years ago
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