Question

Can someone help me find the derivative of

y=(2sqrx +7)^3/ (x^3-3x^2+1)^7

Answer 1

i would use logarithmic differentiation for something so annoying

Answer 2

i don't know what that is. I keep doing it, but i think i am making mistakes some were because part of my answer is wrong

Answer 3

The explanation here should help you:
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/logdiffdirectory/LogDiff.html

Answer 4

but i can't use that on a test, thats why i'm doing these type of questions. They are on my test and I can only use the chain, power, constant and quotient rule

Answer 5

\[y= (2\sqrt{x}+7)^{3}//(x ^{3}-3x ^{2}+1)^{7} \]

Answer 6

Using the quotient rule and the chain rule gives:
\[\frac{\frac{3}{\sqrt{x}}(x ^{3}-3x ^{2}+1)(2\sqrt{x}+7)^{2}-7(2\sqrt{x}+7)^{3}(x ^{3}-3x ^{2}+1)^{6}(3x ^{2}-6x)}{(x ^{3}-3x ^{2}+1)^{14}}\]

Answer 7

ok, thats what I got for the first step when I did it

Answer 8

go on

Answer 9

which can be simplified to:
\[\frac{(2\sqrt{x}+7)^{2}[\frac{3}{\sqrt{x}}-7(2\sqrt{x}+7)(x ^{3}-3x ^{2}+1)^{5}(3x ^{2}-6x)]}{(x ^{3}-3x ^{2}+1)^{13}}\]

Answer 10

I factored out (x^3-3x^2+1)^6

Answer 11

because in the first step don't you have to multiply the derivative of the top by whats on the bottom, so you will have (x^3-3x^2+1)^7. then you factor (x^3-3x^2+1)^6 out of the whole equation .

Answer 12

The quotient rule can be written as:
\[if\ \ y=\frac{u}{v}\ then\ \ \frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v ^{2}}\]
In your question
\[u=(2\sqrt{x}+7)^{3}\]
\[v=(x ^{3}-3x ^{2}+1)^{7}\]
The expression\[(x ^{3}-3x ^{2}+1)^{7}\]
does not arise as a result of the process of differentiation.

Answer 13

the way that I learned it is that f(x)=h(x)/g(x) so the derivative will have the form f'(x)=h'(x)g(x)-g'(x)h(x) divided by g(x)^2

Answer 14

Another way of expressing the quotient rule is as follows:
\[(\frac{f}{g})'=\frac{g.f'-f.g'}{g ^{2}}\]

Answer 15

yea, thats the way i'm using

Answer 16

It amounts to the same as the rule expressed interms of u and v.

Answer 17

ok. but can you use the way that i'm using so I can understand it better. Please. I did this question 3 times and I'm still getting weird answers.

Answer 18

Can you post your initial step here?

Answer 19

sure . Do you want the steps that I did, up to were I'm stuck?

Answer 20

Please post as much as possible.

Answer 21

\[\frac{(3x ^{-1/2})(2\sqrt{x}+7)^{2}(x ^{3}-3x^{2}+1)^{7}-7(x ^{3}-3x ^{2}+1)^{6}(3x ^{2}-6x)(2\sqrt{x}+7)^{3} }{ (x ^{3}-3x ^{2}+1)^{14} }\]

Answer 22

\[\frac{(x ^{3}-3x ^{2}+1)^{6}(2\sqrt{x}+7)^{2}[(3x ^{-1/2})(x ^{3}-3x ^{2}+1)-7(3x ^{2}-6x)(2\sqrt{x}+7)]}{ (x ^{3}-3x ^{2}+1)^{14} }\]

Answer 23

\[\frac{(2\sqrt{x}+7)^{2}[(3x ^{-1/2})(x ^{3}-3x ^{2}+1)-7(3x ^{2}-6x)(2\sqrt{x}+7)] }{ (x ^{3}-3x ^{2}+1)^{8} }\]

Answer 24

thats what I have so far

Answer 25

All your working is correct. I made a typo in my initial differentiation and carried the error through. Sorry for that :(

Answer 26

its ok. But the problem is that I don't know what to do after my third step.

Answer 27

It appears you have an answer to the question and are trying to simplify your result obtained so far to match the answer. Is my understanding of your situation correct?

Answer 28

yes

Answer 29

Can you please post the answer so I can see what the target solution is?

Answer 30

\[-\frac{ 3(2\sqrt{x}+7)^{2}(13x ^{3}-25^{2}-1+49x^{5/2}-98x ^{3/2}) }{ (x ^{3}-3x ^{2}+1)^{8}\sqrt{x} }\]

Answer 31

Looking at the numerator of what you have so far:
\[(2\sqrt{x}+7)^{2}[(3x ^{-\frac{1}{2}})(x ^{3}-3x ^{2}+1)-7(3x ^{2}-6x)(2\sqrt{x}+7)]\]
\[(3x ^{2}-6x)=3x(x-2)\]
Also the following expression can be taken out as a factor of the two terms inside the square brackets giving:
\[(2\sqrt{x}+7)^{2}(3x ^{-\frac{1}{2}})[(x ^{3}-3x ^{2}+1)-(\frac{7}{3x ^{-\frac{1}{2}}})(3x)(2\sqrt{x}+7)(x-2)]\]

Answer 32

This can be further simplified. Please wait.

Answer 33

Continuing with the numerator:
When the second expression in the square brackets is multiplied out we get:
\[14x ^{3}+49x ^{\frac{5}{2}}-28x ^{2}-98x ^{\frac{3}{2}}\]
So the whole expression inside the square brackets becomes:
\[-13x ^{3}+25x ^{2}+1-49x ^{\frac{5}{2}}+98x ^{\frac{3}{2}}\]
If you take out -1 as a common factor of the above expression and include the other terms you will reach the given answer.

Answer 34

so my answer was right. Could i just leave it like my answer and not do what you did

Answer 35

Its difficult to say for sure. But in my opinion most people marking would give aound 8.5 out of 10 for your doing as much as you have done so far and 10 out of 10 for the extra effort going into the simplification.

Answer 36

oh ok

Answer 37

thank you so much

Answer 38

You're welcome :)