Differentiation

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July 3rd 2008, 02:47 AM
dankelly07

Differentiation

Can someone help me revise how to do these...

x^4cos(x^2)

tan(x)/1+x^3 - quotient rule

sin(exp(1+x^2)) - chain rule?
July 3rd 2008, 03:27 AM
Isomorphism

Quote:

Originally Posted by dankelly07

Can someone help me revise how to do these...

x^4cos(x^2)

tan(x)/1+x^3 - quotient rule

sin(exp(1+x^2)) - chain rule?

1) $x^4\cos x^2$

Product Rule: $(x^4\cos x^2)' = (x^4)'\cos x^2 + x^4(\cos x^2)'$

Apply chain rule on $\cos x^2$

$= 4x^3\cos x^2+ x^4 (2x\cos x^2)$

$= 4x^3\cos x^2 + 2x^5\cos x^2$

2) $\frac{\tan x}{1+x^3}$

Yes try quotient rule, and tell us what you get.

3) $z = \sin e^{1+x^2}$

Yes chain rule :)

First sub $u = 1+x^2, y = e^u$ , then $z = \sin e^{1+x^2} = \sin y$

$\frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{du}\frac{du}{dx}$

So $\frac{dz}{dy} = \cos y, \frac{dy}{du} = e^u, \frac{du}{dx} = 2x$ ,

$\frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{du}\frac{du}{dx} = (\cos y)(e^u)(2x) = 2x e^{1+x^2} \sin e^{1+x^2}$
July 3rd 2008, 04:34 AM
colby2152

Quote:

Originally Posted by Isomorphism

1) $x^4\cos x^2$

Product Rule: $(x^4\cos x^2)' = (x^4)'\cos x^2 + x^4(\cos x^2)'$

Apply chain rule on $\cos x^2$

$= 4x^3\cos x^2+ x^4 (2x\cos x^2)$

$= 4x^3\cos x^2 + 2x^5\cos x^2$

Iso, you forgot to derive cosine itself...

$(x^4\cos x^2)' = (x^4)'\cos x^2 + x^4(\cos x^2)'$

$= 4x^3\cos x^2+- x^4 (2x\sin x^2)$

$= 4x^3\cos x^2+- 2x^5\sin x^2)$
July 3rd 2008, 01:41 PM
dankelly07

$ \begin{gathered} u = \tan (x) \hfill \\ v = 1 + x^3 \hfill \\ \hfill \\ \frac{{du}} {{dx}} = \sec ^2 (x) \hfill \\ \frac{{dv}} {{dx}} = 3x^2 \hfill \\ \end{gathered} $
$ \begin{gathered} so \hfill \\ \frac{{3x^2 \tan (x) - 1 + x^3 \sec ^2 (x)}} {{(1 + x^3 )^2 }} ? \hfill \\ \hfill \\ \end{gathered} $
July 3rd 2008, 05:23 PM
Reckoner

Quote:

Originally Posted by dankelly07

$ \begin{gathered} u = \tan (x) \hfill \\ v = 1 + x^3 \hfill \\ \hfill \\ \frac{{du}} {{dx}} = \sec ^2 (x) \hfill \\ \frac{{dv}} {{dx}} = 3x^2 \hfill \\ \end{gathered} $
$ \begin{gathered} so \hfill \\ \frac{{3x^2 \tan (x) - 1 + x^3 \sec ^2 (x)}} {{(1 + x^3 )^2 }} ? \hfill \\ \hfill \\ \end{gathered} $

No. Two problems: first, the quotient rule is $\left(\frac uv\right)' = \frac{vu' - uv'}{v^2}$ ; your numerator is backwards. Second, you did not properly distribute the $\sec^2x$ over the $1 + x^3$ factor (or did you forget to type the parentheses?)

You should get:

$\frac d{dx}\left[\frac{\tan x}{1+x^3}\right]=\frac{\left(1 + x^3\right)\sec^2x - 3x^2\tan x}{\left(1 + x^3\right)^2}$
August 13th 2008, 12:36 AM
dankelly07

hi just looking over this again, could someone just check that example three is correct?

http://www.mathhelpforum.com/math-he...a457bf04-1.gif

http://www.mathhelpforum.com/math-he...3cfe2705-1.gif, then http://www.mathhelpforum.com/math-he...9dcdcf29-1.gif

http://www.mathhelpforum.com/math-he...a1952c08-1.gif

So http://www.mathhelpforum.com/math-he...2ac815e7-1.gif,

I get it up to here..

but below where does sin come from in the last part? should it not be cose^1+x^2...

http://www.mathhelpforum.com/math-he...f7bcda1e-1.gif
August 13th 2008, 01:39 AM
Simplicity

Quote:

Originally Posted by dankelly07

hi just looking over this again, could someone just check that example three is correct?

http://www.mathhelpforum.com/math-he...a457bf04-1.gif

http://www.mathhelpforum.com/math-he...3cfe2705-1.gif, then http://www.mathhelpforum.com/math-he...9dcdcf29-1.gif

http://www.mathhelpforum.com/math-he...a1952c08-1.gif

So http://www.mathhelpforum.com/math-he...2ac815e7-1.gif,

I get it up to here..

but below where does sin come from in the last part? should it not be cose^1+x^2...

http://www.mathhelpforum.com/math-he...f7bcda1e-1.gif

Yes, it should be $2xe^{1+x^2}\cos (e^{1+x^2})$ . See below:
__________________

$z=\sin e^{1+x^2}$

$\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{\mathrm{d}z}{\mathrm{d}u}.\frac{\mathrm{d}u} {\mathrm{d}x}$

Let $u=e^{1+x^2}$ $\therefore z=\sin u$ .

$\frac{\mathrm{d}z}{\mathrm{d}x} = \frac{\mathrm{d}(\sin u)}{\mathrm{d}u}.\frac{\mathrm{d}(e^{1+x^2})}{\mat hrm{d}x}$

$=\cos u . \frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x} = \cos (e^{1+x^2}) . \frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x}$

For $\frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x}$ , use chain rule with $v=1+x^2$

$\therefore \frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x} = 2xe^v = 2xe^{1+x^2}$

$\frac{\mathrm{d}z}{\mathrm{d}x} = 2xe^{1+x^2}\cos (e^{1+x^2})$
August 13th 2008, 04:22 PM
ibnashraf

hey ppl ..... just joined this forum a lil while ago and i'm impressed by what i've seen so far. although i havent done calculus since 5 years ago (so i'm really really rusty! ) but just looking at the recent posts brings back some wonderful memories.

amm .... i'm new to all of this but can someone tell me how to get the different mathematical symbols when posting? i mean like how to get the dy/dx or integral sign, exponent, fractions etc. is there a clipboard on the forum with the symbols on it or are they from an external program or what?

thanks (Hi)
August 14th 2008, 11:50 PM
Prove It

Quote:

Originally Posted by ibnashraf

hey ppl ..... just joined this forum a lil while ago and i'm impressed by what i've seen so far. although i havent done calculus since 5 years ago (so i'm really really rusty! ) but just looking at the recent posts brings back some wonderful memories.

amm .... i'm new to all of this but can someone tell me how to get the different mathematical symbols when posting? i mean like how to get the dy/dx or integral sign, exponent, fractions etc. is there a clipboard on the forum with the symbols on it or are they from an external program or what?

thanks (Hi)

It's done using LaTeX

In Square brackets write math, write a mathematical expression using LaTeX code, then write /math in square brackets.

For help check the LaTeX sub forum...