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

ganesh
Administrator
Registered: 2005-06-28
Posts: 24,247

### 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)

.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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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

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

ganesh
Administrator
Registered: 2005-06-28
Posts: 24,247

### Re: Differential Calculus

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

$3\left(3x^{2}+4x-5\right)^{2} \cdot \left(6x+4\right)$

This would produce:

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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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
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Differential Calculus

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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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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