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Calculus help pleaseeeee :(
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  #1  
Old 10-24-2010, 10:30 PM
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Default Calculus help pleaseeeee :(

Find dy/dx if y=(x^2+5)^10

Last edited by i Maax i : 10-25-2010 at 12:13 AM.
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  #2  
Old 10-24-2010, 11:09 PM
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Default Re: Calculus help pleaseeeee :(

Did you learn product rule? That's what you need to find the derivative of sinxcosx. I'll paste over what I put in the other guys thread.

Product rule is

y = uv
y' = u'v + uv'

Say y = xlnx

y' = x 1/x + (1)lnx
y' = lnx + 1


Now that you have an example, try it again and I'll see if it is correct.

For F'(x), and you want to say find when x=0, just plug it in. F'(0) = .... whereever there are x's plug in 0 and solve it. You have the 2nd part right there but the derivative isn't. Just to answer it, sin pi/3 cos pi/3 is solvable, you can draw a triangle out to demonstrate it. Remember pi = 180 degrees so pi/3 is 60. Sin 60 is rt3 / 2.
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Old 10-24-2010, 11:18 PM
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Default Re: Calculus help pleaseeeee :(

Yeah I learned the product rule. The pi/3 is messing me up.

How'd you know that sin60 is √3/2 ?

Here's what I found:

F'(pi/3) = F'(sin(pi/3) cos(pi/3)) = (cos(pi/3))(-sin(pi/3)) = (√3/2)(-1/2)

Is that good?
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Old 10-24-2010, 11:30 PM
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Default Re: Calculus help pleaseeeee :(

You didn't do the product rule correctly. You should have gotten F'(x)=sin(x)*-sin(x)+cos(x)*cos(x). Look at koot's post for clarification on how I got this.

The sin of pi/3=sqrt(3)/2, and the cos of pi/3=1/2. It's just something that you memorize.

Take a look at this for help:

Your answer should be -1/2
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Old 10-24-2010, 11:33 PM
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Default Re: Calculus help pleaseeeee :(

Yeah I just printed something similar and my book says -1/2 too.

I'm going to try again I think I know where I messed up; I forgot to do the product rule before derivating my fontion.
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Old 10-24-2010, 11:48 PM
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Default Re: Calculus help pleaseeeee :(

Quote:
Originally Posted by i Maax i View Post
So... if:
F(x)= sinx cosx

Find:
F'(x)
F'(pi/3)

Second:
F'(pi/3)= F'(sin(pi/3) cos(pi/3)) = I'm stuck there.
First:

F(x)=sinxcosx
F'(x)= -(sinx)^2+(cosx)^2
F'(pi/3)= -(sqrt(3)/2)^2+(1/2)^2 = -(3/4)+(1/4) = -2/4 = -1/2
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  #7  
Old 10-24-2010, 11:49 PM
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Default Re: Calculus help pleaseeeee :(

Quote:
Originally Posted by DtheK View Post
First:

F(x)=sinxcosx
F'(x)= -(sinx)^2+(cosx)^2
F'(pi/3)= -(sqrt(3)/2)^2+(1/2)^2 = -(3/4)+(1/4) = -2/4 = -1/2
Ya thanks, I finally arrived to the answer after realizing I forgot to put my god damn calculator in RAD mode..........................................
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