thinay Posts: 40, Reputation: 2 Junior Member #1 Jul 29, 2009, 11:45 PM
Derivatives (differentitation of algebraic function)
hello. Can anyone please help me in these problems.. :) Our prof gave us a long test a day ago.. but sad to say I failed to answer the 4 items in that test. Our prof gave the answers after the test but there are no solutions included. So, please help me with the solutions. Find the derivative.

1.) square root of 9-x^2

2.) cube root of (3x+1)^4
Answer: 4 cube root of 3x+1

3.) x square root of 1-x
Answer: (2-3x)/(2 square root of 1-x)

4.) square root of ((1-x)/(1+x))

As you can notice, they all have square root. And I admit I'm weak when it comes with that part. Hope you can help me. Thanks in advance! :)
 KISS Posts: 12,510, Reputation: 839 Uber Member #2 Jul 30, 2009, 12:55 AM

square root of 9-x^2

sqrt is 1/2 power, cube root is 1/3 power etc.

Rewriting:

(9-x^2)^(1/2)

1/2 (9-x^2)^ (1/2-1)* -2x ; See explanation

-x (9-x^2)^(-1/2)

http://www.ucl.ac.uk/Mathematics/geo...wsnb/pow5.html

-x/(9-x^2)^(1/2); use 1/ rule

Explanation:

OK, the 1/2 is the exponent or sqrt rewritten)

The (9-x^2) is the next part of the dervitive eqn.

(1/2-1) is raising it to the power reduced by 1

and -2x is the derivative of (9-x^2)

Now see what you can do to the rest of them.
 thinay Posts: 40, Reputation: 2 Junior Member #3 Jul 30, 2009, 05:28 AM

when I tried solving the 4 items, Ive got the correct answer in #s 1 and 2. But in #s 3 and 4, I end up with wrong answers. :( I think my solutions are wrong.

Here's my answer that I got:
3.) -1/(2 sqrt of 1-x)
4.) -1/(1+x)

 galactus Posts: 2,271, Reputation: 282 Ultra Member #4 Jul 30, 2009, 05:59 AM
4.)$\sqrt{\frac{1-x}{1+x}}$

As you can notice, they all have square root. And I admit I'm weak when it comes with that part. Hope you can help me. Thanks in advance! :)
This involves the chain rule and quotient(or product) rule.

The chain rule is basically the derivative of the inside times the derivative of the outside.

Derivative of outside:

$\frac{1}{2}(\frac{1-x}{1+x})^{\frac{-1}{2}}$

Derivative of inside using the quotient rule:

$\frac{(1+x)(-1)-(1-x)(1)}{(1+x)^{2}}=\frac{-2}{(x+1)^{2}}$

So, upon multiplying, we get:

$\frac{1}{2}(\frac{1-x}{1+x})^{\frac{-1}{2}}\cdot \frac{-2}{(x+1)^{2}}$

$\Huge\fbox{\frac{-1}{(x+1)^{2}\sqrt{\frac{1-x}{1+x}}}}$

That's it. Not bad, huh? Now, you attempt the other one.
 thinay Posts: 40, Reputation: 2 Junior Member #5 Jul 30, 2009, 06:28 AM

... but according to the book where we take our test the answer in number 4 is this:
-1/(1+x)(sqrt of 1-x^2)
 galactus Posts: 2,271, Reputation: 282 Ultra Member #6 Jul 30, 2009, 10:00 AM
It's the same thing. To get your form, multiply top and bottom by $(1+x)$

Like so, $\frac{-1}{(x+1)^{2}\sqrt{\frac{1-x}{1+x}}}\cdot \frac{x+1}{x+1}$

$\frac{-\sout{(x+1)}}{(x+1)^{\not{2}^{1}}\frac{\sqrt{1-x}}{\sqrt{1+x}}\cdot (1+x)}$

$\frac{-1}{(x+1)\sqrt{1-x}\sqrt{1+x}}$

$\frac{-1}{(x+1)\sqrt{(1-x)(1+x)}}$

Remember that $(1-x)(1+x)=1-x^{2}$

$\frac{-1}{(x+1)\sqrt{1-x^{2}}}$
 thinay Posts: 40, Reputation: 2 Junior Member #7 Jul 30, 2009, 04:40 PM

Ah.. I see.. the answer in the book is already simplified. Tnx! :)
 galactus Posts: 2,271, Reputation: 282 Ultra Member #8 Jul 30, 2009, 05:58 PM
Yes, often textbooks write their solutions in ways that is unrecognizable by the student.

For #3, use the product rule.

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