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These are from a class homework assignment that will be graded.

lim (sqrt((9x^2)+4))/x+3

x->00

lim (1/2)((9x^2)+4)^(-1/2)(18x) //L'Hopital rule

x->00

from there i got

lim (9x)/(sqrt((9x^2)+4))

x->00

where do i go from here?

lim sqrt(x^2 +6)/4x+7

x->-00

lim ((.5)(x^2 +6)^-1/2 (2x)) / 4

x->-00

where to from here?

lim sinxcosx

x-> (2pi)/3

i got that it equals -.433

for these i need to find the dy/dx

y=(sqrt(6x+3))/((4x-5^3)

i got

y'= (((6x+3)^(-1/2))[(1/2)(4x-5)-12(6x+3)]) / ((4x-5)^4)

y=(sqrt(5x+2))/((3x-1)^5)

i got

y'= (((5x+2)^-.5)[(5/2)(3x-1)-15])/((3x-1)^6)

y= (x^4)cotx

i got

y'= (-x^4)cscx + (4x^3)cotx

y=(3x^2)cscx

i got

y'=(-3x^2)cscxcotx + 6xcscx

some other weird problems:

find f'(3) using the alternate form of a derivative if f(x) is | x-3 |

well.....all i know is f'(c)= lim as x -> c is (f(x)-f(c))/x-c

but do i need to break it down into piece-wise stuff to solve this one?

y=sin3x on [0, pi/6]

well...

this is mean value theorem so, i solved for both endpoints and got the dy/dx of the function and set it equal to the slope of the secant line...im suck on the step where i set it equal...so i have:

(3(sqrt(3)))= cos(2x)

i believe i need to use cos double angle theorem or something but i'm not sure how that would work.

any help on these would be GREATLY appreciated, this hw is not due until friday so I'm not last minuting it or anything...been working on a giant packet of this stuff for 3 weeks UGH...its all fun though :D