# Differentiation

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• Jul 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?
• Jul 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) $\displaystyle x^4\cos x^2$

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

Apply chain rule on $\displaystyle \cos x^2$

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

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

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

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

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

Yes chain rule :)

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

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

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

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

Originally Posted by Isomorphism
1) $\displaystyle x^4\cos x^2$

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

Apply chain rule on $\displaystyle \cos x^2$

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

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

Iso, you forgot to derive cosine itself...

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

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

$\displaystyle = 4x^3\cos x^2+- 2x^5\sin x^2)$
• Jul 3rd 2008, 01:41 PM
dankelly07
$\displaystyle \begin{gathered} u = \tan (x) \hfill \\ v = 1 + x^3 \hfill \\ \hfill \\ \frac{{du}} {{dx}} = \sec ^2 (x) \hfill \\ \frac{{dv}} {{dx}} = 3x^2 \hfill \\ \end{gathered}$
$\displaystyle \begin{gathered} so \hfill \\ \frac{{3x^2 \tan (x) - 1 + x^3 \sec ^2 (x)}} {{(1 + x^3 )^2 }} ? \hfill \\ \hfill \\ \end{gathered}$
• Jul 3rd 2008, 05:23 PM
Reckoner
Quote:

Originally Posted by dankelly07
$\displaystyle \begin{gathered} u = \tan (x) \hfill \\ v = 1 + x^3 \hfill \\ \hfill \\ \frac{{du}} {{dx}} = \sec ^2 (x) \hfill \\ \frac{{dv}} {{dx}} = 3x^2 \hfill \\ \end{gathered}$
$\displaystyle \begin{gathered} so \hfill \\ \frac{{3x^2 \tan (x) - 1 + x^3 \sec ^2 (x)}} {{(1 + x^3 )^2 }} ? \hfill \\ \hfill \\ \end{gathered}$

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

You should get:

$\displaystyle \frac d{dx}\left[\frac{\tan x}{1+x^3}\right]=\frac{\left(1 + x^3\right)\sec^2x - 3x^2\tan x}{\left(1 + x^3\right)^2}$
• Aug 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
• Aug 13th 2008, 01:39 AM
Simplicity
Quote:

Originally Posted by dankelly07

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

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

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

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

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

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

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

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

$\displaystyle \frac{\mathrm{d}z}{\mathrm{d}x} = 2xe^{1+x^2}\cos (e^{1+x^2})$
• Aug 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)
• Aug 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...