Wondering if my answers are correct below for the instructions differentiate and simplify..

i)

y=(4x+3)^7/(5x-2)^7

my answer:

y'=[-161(4x+3)^6]/(5x-2)^3

ii)

y=[(7x-2)^0.5]/[(5x+3)^0.5]

y'=(5(10x+10.2))/(2(7x-2)^0.5(5x+3)^2)

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- Aug 9th 2009, 11:47 AMskeske1234differentiate and simplify #2
Wondering if my answers are correct below for the instructions differentiate and simplify..

i)

y=(4x+3)^7/(5x-2)^7

my answer:

y'=[-161(4x+3)^6]/(5x-2)^3

ii)

y=[(7x-2)^0.5]/[(5x+3)^0.5]

y'=(5(10x+10.2))/(2(7x-2)^0.5(5x+3)^2) - Aug 9th 2009, 12:08 PMtom@ballooncalculus
Just in case a picture helps...

http://www.ballooncalculus.org/asy/d...d/quotient.png

As usual, straight continuous lines differentiate downwards with respect to x, and the straight dashed line with respect to the dashed balloon expression. So what you have here is two of the chain rule...

http://www.ballooncalculus.org/asy/chain.png

...wrapped inside the product rule...

http://www.ballooncalculus.org/asy/prod.png

... in this case the right-hand of the two.

Try to fill in the blanks, then simplify the bottom row.

______________________________

Don't integrate - balloontegrate!

Balloon Calculus: worked examples from past papers - Aug 9th 2009, 12:13 PMskeske1234
- Aug 9th 2009, 12:17 PMtom@ballooncalculus
Well, they don't

*look*remotely right... I'll check them again while you study mine?

Hey, simplify mine, then we'll know.

Show us your's, by the way? What did you use, the quotient rule?

For the second one...

http://www.ballooncalculus.org/asy/d.../quotient2.png - Aug 9th 2009, 12:56 PMeXist
Can you show me your work for the first one? I got a completely different answer.

Remember the quotient rule, where u and v are both quantities of x: $\displaystyle \frac{d}{dx} (\frac{u}{v}) = \frac{vu' - uv'}{v^2}$

Also in this case you will be applying the chain rule to both u and v since $\displaystyle u = (4x + 3)^7$ and $\displaystyle v = (5x - 2)^7$ - Aug 9th 2009, 01:23 PMtom@ballooncalculus
Whichever method you choose, it does no harm to know that the quotient rule is only the chain rule wrapped in the product rule, as depicted above. I don't say it's always an overwhelming advantage of the diagram but... please notice that the formula can land you with unnecessarily large powers in the denominator (14) which will probably want to be simplified down to what you already have in the diagram (8).

- Aug 9th 2009, 01:40 PMPlato
Take note that $\displaystyle y = \frac{{\left( {4x + 3} \right)^7 }}

{{\left( {5x - 7} \right)^7 }} = \left( {\frac{{4x + 3}}

{{5x - 7}}} \right)^7 $.

So $\displaystyle y' = 7\left( {\frac{{4x + 3}}

{{5x - 7}}} \right)^6 \left( {\frac{{ - 43}}

{{\left( {5x - 7} \right)^2 }}} \right)$. - Aug 9th 2009, 02:51 PMskeske1234
- Aug 9th 2009, 03:01 PMtom@ballooncalculus
Just in case a picture helps...

... although I must admit (not that one would necessarily care) that using the quotient rule inside the chain rule, while it makes for an easier ride with the formula, doesn't make one nice big picture. Nonetheless...

http://www.ballooncalculus.org/asy/d.../quotient3.png

... into which (in the blank) we can paste the result of...

http://www.ballooncalculus.org/asy/d.../quotient4.png

__________________________________

Don't integrate - balloontegrate!

Balloon Calculus: worked examples from past papers - Aug 9th 2009, 03:08 PMtom@ballooncalculus