OpenStudy (anonymous):

f(x)=1/(4+9x^2) - sketch the region bounded by f, the x-axis, and the line x=1 and x=2

7 years ago
OpenStudy (julie):

hmmm, well, I think basically this is a second derivative function. Paul has some pretty good notes here on how to use the second derivative to glean more about the shape of a function http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtII.aspx

7 years ago
OpenStudy (julie):

seems to explain it well also. but yeah you need information from both the first and second derivatives in order to correctly sketch the region

7 years ago
OpenStudy (julie):

been years since I've done so :p

7 years ago
OpenStudy (julie):

$$f(x)=1/(4+9x ^{2})$$

7 years ago
OpenStudy (julie):

^^ function using the equation editor thingy

7 years ago
OpenStudy (anonymous):

Thanks Julie I think that helps me some!

7 years ago
OpenStudy (julie):

hmmm, honestly going to have to look up the chain rule again lol

7 years ago
OpenStudy (julie):

to even get the first derivative

7 years ago
OpenStudy (julie):

hmmm. does anyone remember the chain rule? (looking)

7 years ago
OpenStudy (shadowfiend):

So we have: $$\frac{1}{4 + 9x^2}$$ Which makes: $$(4 + 9x^2)^{-1}$$ We can apply the chain rule to this by first setting: $$ u = (4 + 9x^2) $$ And: $$\frac{d}{du} u^{-1} = -u^{-2}$$

7 years ago
OpenStudy (shadowfiend):

But the chain rule says that taking the derivative of u when u consists of a function of x means we have to multiply by the derivative of the inner function: $$ \frac{d}{dx} \left(4 + 9x^2\right) = \frac{d}{dx}4 + \frac{d}{dx}9x^2 $$ We can apply basic power rule and such and get: $$ \frac{d}{dx} \left(4 + 9x^2\right) = 0 + 18x = 18x $$ And our end result is: $$ -\left(4 + 9x^2\right)^-2 \cdot 18x = -\frac{18x}{\left(4 + 9x^2\right)^2} $$

7 years ago
OpenStudy (shadowfiend):

That's the first derivative, you can apply the same reasoning to get the second derivative, and then use julie's advice to find the shape of the function.

7 years ago
OpenStudy (shadowfiend):

(Feel free to correct me anyone -- I'm doing this in my head and haven't gotten a chance to double-check myself.)

7 years ago
OpenStudy (anonymous):

One way of remembering the Chain Rule in this type of case is to remember that it will be the derivative of the outside function with the inside function left alone times the derivative of the inside function.

7 years ago
OpenStudy (anonymous):

So, in this case once you write the fucntion as, $$ (4+9x^{2})^{-1}$$ the "outside function" is the outermost function, i.e. the exponent of "-1" while the "inside function" is then $$4+9x^{2}$$, i.e. the stuff on the inside.

7 years ago
OpenStudy (anonymous):

We know that the derivative of say $u^{-1}$ is $-u^{-2}$ so differentiating the outside, while leaving the inside alone would give, $$-\left( 4+9x^{2}\right)^{-2}$$ Then multiply all this by the derivative of the inside, $18x$. Or, putting it all together you get what shadofiend got.

7 years ago
OpenStudy (julie):

wow thanks MD! very helpful. chain rule is a bit... confusing to decipher without explanation

7 years ago
OpenStudy (anonymous):

Yes, it's can be difficult when trying to learn it (or scape the rust off skills long unused :) ). If you recall the outside/inside function way of using the chain rule it can help with a lot of the "simpler" problems. Although I suspect "simpler" is in the eye of the beholder... :)

7 years ago
OpenStudy (anonymous):

okay my text is working again

7 years ago
OpenStudy (julie):

@MD indeed! @mickey indeed!

7 years ago
OpenStudy (shadowfiend):

@MD Thanks -- by the way, for inline LaTeX, I'm afraid you have to use \ ( and \ ) (without the spaces) so that we don't start messing up when people start using $ for money :) So \(u^{-1}\) is \(u^{-2}\) should work.

7 years ago
OpenStudy (anonymous):

Ah, I was wondering about that. It's been ages since I've seriously used LaTeX and I was thinking I'd misremembered how to do it (got a lot of rust there :) ). I'd forgotten about the other way of doing inline LaTeX. Thanks for reminding me! On a side note a preview button might be good when typing into the box directly, i.e without using the Equation editor....

7 years ago
OpenStudy (shadowfiend):

Yep. Ideally we'd like the equation editor itself to allow inline equations.

7 years ago
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