Derivatives and Tangents

**soulmeetsbody** · January 18th 2009, 04:15 PM

Hey, there I have one quick question for now.. I'm in grade 12 calc and we're doing derivatives and have learned the power rule (its a nice shortcut) and i'm having trouble with this one problem:

Write an equation of a tangent to each of the curves at the given point:

$y= \sqrt{3x^3}$ P(3,9)
Heres what I did:

$y=(3x^3)^\frac{1}{2}$
Then the derivative:
...Do I use the power rule on the inside brackets first? or distribute the 1/2 outside the brackets??

$y'=(9x^2)^\frac{1}{2}$
$y'=(3x)$

So my equation turns out to be
9x-y-18=0,
but the answer in the back of the book is 9x-2y-9=0
What am I doing wrong?

**Mush** · January 18th 2009, 04:24 PM

Originally Posted by soulmeetsbody

Hey, there I have one quick question for now.. I'm in grade 12 calc and we're doing derivatives and have learned the power rule (its a nice shortcut) and i'm having trouble with this one problem:

Write an equation of a tangent to each of the curves at the given point:

$y= \sqrt{3x^3}$ P(3,9)
Heres what I did:

$y=(3x^3)^\frac{1}{2}$
Then the derivative:
...Do I use the power rule on the inside brackets first? or distribute the 1/2 outside the brackets??

$y'=(9x^2)^\frac{1}{2}$
$y'=(3x)$

So my equation turns out to be
9x-y-18=0,
but the answer in the back of the book is 9x-2y-9=0
What am I doing wrong?

First of all, you should distribute the power!
$\frac{d}{dx} \sqrt{3x^3} = \frac{d}{dx} (3x^3)^{\frac{1}{2}}$

$= \frac{d}{dx} (3^{\frac{1}{2}}x^{\frac{3}{2}})$

$= \frac{d}{dx} (\sqrt{3}x^{\frac{3}{2}})$

$= \sqrt{3} \frac{d}{dx} x^{\frac{3}{2}}$

$= \sqrt{3} \times \frac{3}{2} \times x^{\frac{1}{2}}$

$= \frac{3\sqrt{3}}{2} x^{\frac{1}{2}}$

$= \frac{3\sqrt{3}}{2}\sqrt{x}$

**soulmeetsbody** · January 18th 2009, 04:34 PM

Thank you very much! It makes a lot more sense now

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