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#1 2009-03-03 02:46:32

ganesh
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Differential Calculus

1. Differentiate the following with respect to x.

(i)



(ii)


(iii)


2. Differentiate the following with respect to x.

(i)


(ii)


(iii)


3. Differentiate the following with respect to x.

(i)


(ii)


(iii)


(iv)


(v)


(vi)
.


Character is who you are when no one is looking.
 

#2 2009-03-10 19:02:26

lady einstein
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Re: Differential Calculus

#2 i)
3 (3x^2 + 4x - 5)^2  *(6x + 4)

i beg to be pardoned for how i express my equations smile

 

#3 2009-03-11 01:17:06

ganesh
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Re: Differential Calculus

lady einstein,
Your answer is perfectly right!
The answer can be expressed in LaTeX by using the following code:-

Code:

[math]3\left(3x^{2}+4x-5\right)^{2} \cdot \left(6x+4\right)[/math]

This would produce:


Character is who you are when no one is looking.
 

#4 2009-06-25 18:36:18

glenn101
Full Member

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Re: Differential Calculus

1.(i) e^x
    -----
     cos(x)

Using Quotient Rule:
Let u=e^x, v=cos(x)
     du/dx=e^x, dv/dx=-sin(x)

dy/dx=   v(du/dx)-u(dv/dx)
               --------------------
                           cos^2(x)
         =    cosx(e^x)-e^x(-sin(x))
               ----------------------------
                            cos^2(x)
     

              =       -e^x(-cos(x)+sin(x))
                   -------------------------
                             cos^2(x)


(ii)    log(x)
      ------------
        sin(x)
Using Quotient Rule:
Let u=log(x), v=sin(x)
     du/dx= 1/x,  dv/dx=cos(x)

dy/dx=       v(du/dx)-u(dv/dx)
                  ---------------------
                            v^2
        =        sinx(1/x)-logx(cos(x))
                   -------------------------
                                sin^2(x)

         =        sin(x)/x-logx(cos(x))
                    ---------------------------
                            sin^2(x)
2. (i) y=(3x^2+4x-5)^3
Using Chain Rule:
Let u = 3x^2+4x-5, y=u^3
     du/dx=6x+4       dy/du= 3u^2

      dy/dx= 3(3x^2+4x-5)^2*(6x+4)



    (ii) y=e^3x^2+2x+3
         dy/dx=6x+2e^3x^2+2x+3
3. (i) y=e^2xcos(3x)
Using Product Rule:
Let u = e^2x,  v=cos(3x)
      du/dx=2e^2x, dv/dx=-3sin(3x)

dy/dx= -e^2x(3sin(3x)+2e^2x(cos(3x))

           
I love Calculus:D
Also, just a suggestion Ganesh.
Could you perhaps make exercises for Integration?

Last edited by glenn101 (2009-06-25 18:39:45)


"If your going through hell, keep going."
 

#5 2009-06-26 07:05:53

bobbym
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Re: Differential Calculus


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#6 2009-11-24 18:08:33

Denominator
Full Member

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Re: Differential Calculus

I think 2 ii is
(6x+2)e^(3x^2+2x+3)

 

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