Calculus help

Test Preparation
1

<p>Hey.. I'd appreciate it if you could guys could verify my answers for the following questions..</p>

<p>Find dy/dx:</p>

<p>1) y=arcsin(2x)
y'= (2)/sqrt(1-4x^2)</p>

<p>2) y=arctan(x^2)
y'= 2x/(1+x^4)</p>

<p>3) 3sqrt(x)=sqrt(y)
dy/dx=( 3.(x)^(-.5) / (y)^(-.5))</p>

<p>4) xy+y=x
dy/dx = (-y)/(x+y)</p>

<p>5) e^cosy=x^3 arctan y</p>

<p>(e^cosy) (-siny. dy/dx ) = 3x^2 + (1/1+y^2)dy/dx</p>

<p>6) Find slope of the following when x equals the indicated value.</p>

<p>A) xy+x+y=8, x=2</p>

<p>dy/dx= (-y-1)/(x+Y)
dy/dx= -3/4</p>

<p>B) sin^2y + cos^2x=4, x=(pi/2)</p>

<p>9) 2x^2+y^2=4
Dy/dx= -2x/y</p>

<p>10) ln(xy)=4
ln ( x. dy/dx + y ) = 0
1/ x.dy/dx + y = 0
dunno how to proceed
dunno how to do this..</p>

<p>thanks much</p>

2

<p>not sure about your answer for 3
i get 3sqrt(y/x), which is equal to 9</p>

3

<p>how'd you get a single value though since the value of y and x is not specified? so its probably gonna hve an x or y in the answer..</p>

<p>thanks much for your help.. I"d appreciate it if someone could verify the other answers too.</p>

4

<p>well, if dy/dx = 3sqrt(y/x) is correct... you can plug 3sqrt(x) in for sqrt(y) because that's given. this way you get 9sqrt(x)/sqrt(x) which is just 9</p>

5

<p>your answer to 4 is wrong. the actual answer is dy/dx = (1-y)/(x+1), which is equal to (x+1)^(-2). solve for y and use quotient rule.</p>

<p>you forgot product rule on 5, unless the right side was actually
x^3 + arctan y</p>

6

<p>For 6A I got:</p>

<p>xy + x + y = 8
y + xy' + 1 + y' = 0
y + 2y' + 1 + y' = 0
3y' = -y-1
y' = -1/3(y+1)</p>

7

<p>haha i messed up. edited version:</p>

<p>6B) sin^2y + cos^2x=4, x=(pi/2)
(2sin(y))(cos(y))(dy/dx) + (2cos(x))(-sin(x)) = 0
solving for dy/dx:
dy/dx = (2sinxcosx)/(2sinycosy)</p>

<p>there's a trig identity that sin(2A) = 2sinAcosA. therefore,
dy/dx = (sin(2x))/sin(2y)</p>

<p>so we need to know y to find the slope. plugging pi/2 in for x in the original equation, we get y = 2 or -2</p>

<p>so dy/dx = (sin(pi)) / sin(4)
dy/dx = 0/sin(4)
dy/dx = 0</p>

8

<p>I agree with your solution for #9</p>

9

<p>ok time to do #10</p>

<p>10) ln(xy)=4
using chain rule and product rule...
(1/(xy)) (y + x (dy/dx)) = 0
1/(xy) isn't going to be 0, so the other term has to
y + x(dy/dx) = 0
dy/dx = -y/x</p>

<p>from the original we know that xy = e^4, so y = (e^4)/x</p>

<p>dy/dx = -(e^4)/(x^2)</p>

10

<p>Hmm I got 0 for 6B
sin^2y + cos^2x=4, x=(pi/2)</p>

<p>sin2 y + cos2 x = 4
2y'sinycosy - 2cosxsinx = 0
2y'sinycosy = 2cosxsinx
y' = cosxsinx/(cosysiny)
cos(pi/2) = 0, so the answer is 0</p>

11

<p>I agree with your solutions to #s 1 and 2.</p>

12

<p>yeah i like your answer for that one neo</p>

13

<p>9) 2x^2+y^2=4</p>

<p>4x + 2y(dy/dx) = 0
2y(dy/dx) = -4x
dy/dx = -2x/y</p>

<p>hey you're absolutely right</p>

14

<p>
[quote]

hey you're absolutely right

[/quote]
</p>

<p>sweet sarcasm :)... having probs with these kinda questions doh.. need to practice more..</p>

<p>Thanks much guys.. appreciate it</p>

15

<p>no problem</p>

<p>btw no sarcasm there, you were right on #9</p>

16

<p>A quick observation on (3):
If you square both sides, you get 9x = y, from which dy/dx = 9 .</p>

<p>You may also want to rewrite some of your answers so that they are functions of x alone, rather than of x & y.</p>

17

<p>haha wow, i didn't catch that when i did it</p>

18

<p>Please help me
What is the property of sin2theata?</p>

19

<p>sin 2x = 2sinxcosx</p>

20

<p>thanks!
what is the derrivative of: 1/X^1/X?</p>