By the def
lim f(x + h ) - f(x)
/h
Plug in (x+h) into f
lim (x+h)^2 + 5(x +h) - (x^2 + 5x)
/h
Foil
lim x^2 + 2xh + h^2 + 5x + 5h - x^2 - 5x
/h
Cancel like terms
lim 2xh + h^2 + 5h
/h
Reduce everything by h
lim 2x + h + 5
Take the limit and
Plug in 0 for h
2x + 5
2006-06-30 03:45:13
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answer #1
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answered by theFo0t 3
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Derivative of a functions is indicated as "f-prime-of-x":
f ' (x) instead of f (x)
Use the power rule, which means take each term in the expression and subtract one from the power of "x"
So if you have x^2, it becomes x^(2-1) = x^1 =x
Also, multiply the original power of "x" with the coefficient of the term. If you had 3x^2, it would be 2 * 3 * x^1 = 6x.
For f(x) = x^2 + 5x,
2 * x^(2-1) + 1 * 5 * x^(1 - 1)
=2x^1 + 5x^0
=2x + 5 * 1
=2x + 5
2006-07-01 16:14:19
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answer #2
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answered by Anonymous
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2x + 5
2006-06-30 03:39:39
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answer #3
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answered by Anonymous
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2x + 5
2006-06-30 03:39:16
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answer #4
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answered by Bill S 6
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The derivative formula is nx^(n-1)
f(x) = x^2+5x
f'(x) = 2x^(2-1)+5
f'(x) = 2x+5
2006-06-30 07:29:47
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answer #5
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answered by parag1010 3
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2x+5
2006-06-30 03:39:15
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answer #6
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answered by Black Fedora 6
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