2006-09-23 04:38:58 · 6 answers · asked by mspeanut01 1 in Science & Mathematics ➔ Mathematics
Well dear its like your ex-question i explained , i mean same way
but i will tell you what your correct answer is;
f(x) = (2x+11) / (10x-7)
u(x) = 2x + 11 ; u'(x) = 2
v(x) = 10x - 7 ; u'(x) = 10
f'(x) = [2* ( 10x - 7) - 10*( 2x+11)] / [( 10x - 7)^2]
f'(x) = [2/ ( 10x - 7)] - [10*( 2x+11)] / [( 10x - 7)^2]
Good Luck dear.
2006-09-23 06:32:27 · answer #1 · answered by sweetie 5 · 1⤊ 0⤋
Use the quotient derivative rule. In the textbook we use, it is described with this little poem:
Low dee high minus high dee low over the square of what's below.
(low or what's below means denominator, high means numerator, dee means derivative)
That translates to
[(10x-7)(2)-(2x+11)(10)]/
[(10x-7)^2]
(20x-14-20x-110)/
[(10x-7)^2]
-124/
[(10x-7)^2]
Hope that's useful.
2006-09-23 04:54:27 · answer #2 · answered by just♪wondering 7 · 0⤊ 1⤋
((10x-7)(2) - (2x+11)(10)) / (10x-7)^2
d/dx f(x)/g(x) = (g(x)f'(x) - f(x)g'(x)) / g(x)^2
If you have a calculus book with you, there HAS TO BE a bunch of derivative tables in there.
2006-09-23 04:53:09 · answer #3 · answered by عبد الله (ドラゴン) 5 · 0⤊ 1⤋
f(x) = (2x + 11)/(10x - 7)
f'(x) = (((2x + 11)'(10x - 7)) - ((2x + 11)(10x - 7)'))/((10x - 7)^2)
f'(x) = (2(10x - 7) - (10(2x + 11)))/((10x - 7)(10x - 7))
f'(x) = (20x - 14 - 20x - 110)/(100x^2 - 70x - 70x + 49)
f'(x) = (-124)/(100x^2 - 140x + 49)
2006-09-23 07:18:39 · answer #4 · answered by Sherman81 6 · 0⤊ 1⤋
-124/(10x-7)^2
2006-09-23 04:49:38 · answer #5 · answered by Greg G 5 · 0⤊ 1⤋
[(10x-7)*(2)-(2x+11)*(10)]/(10x-7)^2
2006-09-23 04:43:46 · answer #6 · answered by bruinfan 7 · 0⤊ 1⤋