ok i get that now, but when i try to isolate the dy/dx the big bracket is too hard to get the (dy/dx) out of..
this is what i got when i expanded it.
[{(3x^3)+(3x^2)+(x)(dy/dx)+(dy/dx)-(x^3)-(y)} / (x+1)^2 ]
then only 2 terms in then can be groups the 3x^3 - x^3 and i don't know where...
so it would look this so:
-sin(xy^2)[(y)^2+(2y)(dy/dx)(x)] - [ {(3x^2)(x+1)(dy/dx)-(x^3+y)} / (x+1)^2] = sec(x)tan(x)sin(y) + cos(y)sec(y)(dy/dx)
so many dy/dx 's should i still try and bring them all on one side?
assistance asap is needed !
Using implicit differentiation, find dy/dx given that:
cos(xy^2) - (x^3+y) / (x+1) = sec(x)sin(y)
i am horrid at these, i came up with a few lines, please check them
-sin(xy^2)[(y)^2+(2y)(dy/dx)(x)] - [ {(3x^2)(x+1)-(x^3+y)} / (x+1)^2] = sec(x)tan(x)sin(y) +...
okay and for the first one you are saying that i should break it up into 2 parts
the first being:
lim x \rightarrow x^-
so i would get x/-x
=-1
and then:
lim x -> x
gettinng x/x
= 1
? is this what you are referring too?
hey sry i forgot the second part
Question #2: Find the point on the curve y=(x+1) ³(1-3x) ² where the slope of the tagent line is horizontal
sorry about that
Hey i have 2 questions here that i have finished and i have no answers to my solutions so i would like someone to check it over and see where i went wrong , I am not 100% sure on this stuff and need some help , thanks a lot !
Question 1: Using the definition of a derivative, show that ƒ (x) =...
ok so for part a) i siad that cause when i plotted my points it started at x=2 , so that's why i had my domain as [2,\infty]
and for the rang it was the same thing, i just looked at my graph and tried to do it...
and also the same thing for b).. um I am looking at them again and i can't...
Hey here is the question has a few parts to it.
Let f(x)= (x-1)^1/2 and g(x)= 2x^2/x^2+1
a) State the domain and range of f using interval notation.
b) State the domain and range of g using interval notation. (Hint: At
x=0, g is 0. For x(cannot=0) it helps to write g(x)=2/1+1/x^2 and...
okay thank you very much i have this solved , i broke the horizontal and vertical up, then isolated for the m3 , then equated them to get rid of the m3.
i got:
-38 / 30cos(Theta) = -65.8 / 30sin(theta)
cross multiplied and got 30 degress as my angle. :D
Advanced Momentum Question -- need help
The Question
A bomb initially at rest on a smooth, horizontal surface explodes into 3 pieces. Two pieces fly across the surface at a 60 degree angle to each other: a 2.0 kg at 20 m/s and a 3.0 kg piece at 12 m/s. The 3rd pieces flys across the surface...
Question: You happen to be the one driving down the Highway that this person pulls in front of. You prepare to make a fast lane switch but are boxed in by a transport truck in the fast lane. Thus, you step on the brake pedal. You are traveling 27.777 .. m/s (100 km/h) and are 100 m behind the...
hey i was wondering if this is what we are looking at , cause he said the v=854cosx , but i can't how this works.
http://img503.imageshack.us/img503/6760/223pf.png