# Finding the derivative

• May 11th 2010, 07:28 PM
spoc21
Finding the derivative
Find the derivative of the following function:

f(x) = (2x-3)^2/(x^3-7)^3

I am very confused here. Don't know what rule to use, as the exponents are outside the bracket. I would appreciate any help!

Thanks http://www.mathhelpforum.com/math-he...cons/icon7.gif
• May 11th 2010, 07:32 PM
Chris L T521
Quote:

Originally Posted by spoc21
Find the derivative of the following function:

f(x) = (2x-3)^2/(x^3-7)^3

I am very confused here. Don't know what rule to use, as the exponents are outside the bracket. I would appreciate any help!

Thanks http://www.mathhelpforum.com/math-he...cons/icon7.gif

You will first need to use the quotient rule, and then a combination of power rule and chain rule.

Try to attempt the problem given this information and let us know where you get stuck. :)
• May 11th 2010, 07:34 PM
mr fantastic
Quote:

Originally Posted by spoc21
Find the derivative of the following function:

f(x) = (2x-3)^2/(x^3-7)^3

I am very confused here. Don't know what rule to use, as the exponents are outside the bracket. I would appreciate any help!

Thanks http://www.mathhelpforum.com/math-he...cons/icon7.gif

Use the quotient rule. u = (2x - 3)^2 and v = (x^3 - 7)^3. To differentiate u and v, either expand and differentiate in the usual way, or use the chain rule.
• May 11th 2010, 09:28 PM
spoc21
thanks guys!

So I end up with the following result (am very unsure about it):

Derivative:

$2x-3(4(x^6-32x^3+27x^2+49))$
---------------------------------
$(x-7)^6$

can some one please confirm if this is the correct answer.I would really appreciate it!

Thanks http://www.mathhelpforum.com/math-he...cons/icon7.gif
• May 12th 2010, 12:58 AM
mr fantastic
Quote:

Originally Posted by spoc21
thanks guys!

So I end up with the following result (am very unsure about it):

Derivative:

$2x-3(4(x^6-32x^3+27x^2+49))$
---------------------------------
$(x-7)^6$

can some one please confirm if this is the correct answer.I would really appreciate it!

Thanks http://www.mathhelpforum.com/math-he...cons/icon7.gif

The review process will be much easier and more effective if you post all your working (ie. show every step) rather than just giving a final answer.
• May 12th 2010, 08:57 AM
tom@ballooncalculus
Just in case you're interested (like many people) in avoiding the quotient rule, which can feel like overkill, and always doubles the power of the denominator, sometimes unnecessarily as here... then, how about...

http://www.ballooncalculus.org/asy/diffProd/quot5.png

... where

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

... is the product rule, straight lines differentiating downwards with respect to x. Then you don't need u and v. You still need the chain rule, but you can zoom in on this, for example in the right hand fork...

http://www.ballooncalculus.org/asy/diffProd/quot5a.png

... where...

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

... is the chain rule. Straight continuous lines still differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

Another way of avoiding the quotient rule is logarithmic differentiation - taking logs of both sides of y = the function. Then solve for dy/dx in the bottom row of...

http://www.ballooncalculus.org/asy/diffLog/quot1.png

_________________________________________
Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!
• May 13th 2010, 10:11 AM
spoc21
ok here's the working:

$f(x)=(2x-3)^2/(x^3- 7)^3$
$f' (x)=((x^3-7)^3 [2(2x-3)(2) ]-[3(x^3-7)^2 (3x^2 )])/[(x^3-7)^3 ]^2$
$f' (x)=((x^3-7)^3 [4(2x-3)]-[9x^2 (x^3-7)^2](2x-3)^2)/(x^3-7)^6$
We need to simplify:

$f' (x)=((x^3-7)[4(2x-3) ]-(9x^2)(2x-3)^2)/(x^3-7)^4$
We need to simplify Further:
$f' (x)=(2x-3)[4(x^3-7)-9x^2 (2x-3) ]/(x^3-7)^4$

thats my final answer ^^...I am supposed to represent the derivative in its simplified form..Is my working correct?

Thank you! :)

• May 13th 2010, 02:27 PM
tom@ballooncalculus
Quote:

Originally Posted by spoc21
We need to simplify:

$f' (x)=((x^3-7)[4(2x-3) ]-(9x^2)(2x-3)^2)/(x^3-7)^4$

Good. (The quotient rule had to take you up to power 6 in the denominator, but you're back down again.)

Quote:

We need to simplify Further:
$f' (x)=(2x-3)[4(x^3-7)-9x^2 (2x-3) ]/(x^3-7)^4$

Fine. Why not expand and collect in the square brackets?