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#1 2009-03-02 03:46:32

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,818

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.

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#2 2009-03-09 20:02:26

lady einstein
Member
Registered: 2009-02-17
Posts: 7

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

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#3 2009-03-10 02:17:06

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,818

Re: Differential Calculus

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

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#4 2009-06-24 20:36:18

glenn101
Member
Registered: 2008-04-02
Posts: 108

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 (2009-06-24 20:39:45)


"If your going through hell, keep going."

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#5 2009-06-25 09:05:53

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 88,762

Re: Differential Calculus


In mathematics, you don't understand things. You just get used to them.

I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.

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#6 2009-11-23 19:08:33

Denominator
Member
Registered: 2009-11-23
Posts: 155

Re: Differential Calculus

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

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