Differential Coefficient

**zacky08** · December 2nd 2009, 12:16 PM

Find the differential coefficient for the following different functions to demonstrate the use of function of a function; the product and quotient rules:

a) Y=(1+2x) squared
B) y = 6x squared Sin 2x
c) s= 2x squared
Cos3x

Can you help differentiate it please, thanks.

**pickslides** · December 2nd 2009, 12:27 PM

Hi Zacky

Originally Posted by zacky08

a) Y=(1+2x) squared

Use the chain rule here

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

make $u = 1+2x$ and $y=u^{2}$

Originally Posted by zacky08

B) y = 6x squared Sin 2x

Use the product rule here

$y = u\times v \Rightarrow y' = vu'+uv'$

make $u = 6x^2$ and $v = \sin(2x)$

Originally Posted by zacky08

c) s= 2x squared
Cos3x

Use the quotient rule here

For $y = \frac{u}{v} \Rightarrow y' = \frac{vu'-uv'}{v^2}$

make $u = 2x^2$ and $v=\cos(3x)$

**pickslides** · January 31st 2010, 01:07 PM

Use the chain rule here

make $u = 1+2x$ and $y=u^{2}$

Here's how to start....

First from $u = 1+2x$ find $\frac{du}{dx}$

Spoiler:

then from $y=u^{2}$ find $\frac{dy}{du}$

Spoiler:

After you have these, put them into the RHS of the equation below.

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

**zacky08** · January 31st 2010, 01:22 PM

Thank you so much, could you be so kind to do for the other 2 questions aswell if possible, thanks gracefully, Zacky.

**pickslides** · January 31st 2010, 01:31 PM

Originally Posted by zacky08

Thank you so much, could you be so kind to do for the other 2 questions aswell if possible, thanks gracefully, Zacky.

The idea in the following 2 questions are the same as the previous post. You need to post your attempts on this forum. It is Math 'Help' Forum not Math 'Do' Forum

Also google the product and quotient rules. There are heaps of examples out there.

**zacky08** · January 31st 2010, 02:11 PM

Use the product rule here

make

and

.

Ok firstly, i do 6x² + sin(2x) + sin(2x) + 6x² which gives me sin12x² + sin12x² = 2sin24x4. Am i on the right lines?

**pickslides** · January 31st 2010, 02:18 PM

Originally Posted by zacky08

Ok firstly, i do 6x² + sin(2x) + sin(2x) + 6x² which gives me sin12x² + sin12x² = 2sin24x4. Am i on the right lines?

No this is all types of wrong.

Firstly there is a difference between $u$ and $u'$ . $u'$ is the derivative of $u$

For example if $u = x^2$ then $u' = 2x$

Do you understand this concept?

Also

$<br /> \sin(2x) + \sin(2x) = 2\sin(2x) \neq \sin(12x^2) <br /> <br />$

**zacky08** · January 31st 2010, 02:24 PM

Ah ok, i kind of grasp it now.

So, Sin (2x) + 12x then 12x + Sin (2x) then?

Which would be Sin(24x) + Sin(24x) = 2Sin (48x)

**pickslides** · January 31st 2010, 02:39 PM

No, not even close.

This is the rule.

$y = u\times v \Rightarrow y' = vu'+uv'$

You need to find u' and v' using the following information.

$u = 6x^2$ and $v = \sin(2x)$

I get $u' = 12x$ and $v' = 2\cos(2x)$

The real question is, do you know how I found these?

**zacky08** · January 31st 2010, 02:50 PM

6x² differentiated is 12x, and sin is differentitated into cos and same thing with (2x) to give 2cos(2x). Yeah i realised what i did wrong, still a bit vague. But now that i got

and

, what would be the next step? do i put this into the Dy by Dx again?

**pickslides** · January 31st 2010, 02:58 PM

The answer is $y' = vu'+uv'$

So from your working

$y' = 12x\sin(2x)+12x^2\cos(2x)$

You should also factor out $12x$ giving

$y' = 12x(\sin(2x)+x\cos(2x))$

**zacky08** · January 31st 2010, 03:08 PM

Finally, the Quotient Rule.

make

and

I would get.

Cos (12x) - Cos (12x)
2cos(9x)

**pickslides** · January 31st 2010, 03:10 PM

Originally Posted by zacky08

Finally, the Quotient Rule.

make

and

I would get.

Cos (12x) - Cos (12x)
2cos(9x)

One step at a time there big fella, firstly what did you get for $u'$ and $v'$ ?

**zacky08** · January 31st 2010, 03:14 PM

i got 4x and

i got -sin (9x)

**pickslides** · January 31st 2010, 03:20 PM

$u'$ is correct $v' = -3\sin(3x)$

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