Calculus Question!!! ANY HELP WOULD BE GREATLY APPRECIATED!!!! =)?

Question

Find and correct to two decimal places, the slope of the curve y= x^3 (squareroot of: 3 + 3x^2) (divided by) (3x-4)^2
at the point where x =1. PLEASE HELP ME!!!

Answer 1

You have 3 functions of x, which can be represented as X1(x) * X2(x) * X3(x), where X1(x) is x^3, X2(x) is sqrt(3+3x^2) and X3(x) is (3x-4)^-2. I think you can find dy/dx as
X2*X3*dX1/dx)+X1*X3*dX2/dx+X1*X2*dX3/dx.
This is a bit of dogwork, but the derivitives are not all that bad. Once you clean up the resulting mess, you simply substitute 1 in for x.

Answer 2

yikes!
Here is the derivative
3x^2(3+3x^2)^(1/2)/[(3x-4)^2]+ 3x^4/[(3+3x^2)^(1/2)/(3x-4)^2] - 6x^3(3+3x^2)^(1/2)/[(3x-4)^3]
plug at x=1
19(6)^(1/2)/(2)=23.27

note: the power of 1/2 is "the square root of"

Answer 3

Go look on webmath.com....you MIGHT find the answer...