Differential Calculus midterm

**VX-1** · September 24th 2008, 05:36 PM

Hey guys can you help me with the questions on this Math midterm?

**colby2152** · September 24th 2008, 05:42 PM

What exactly do you need help with? This is one of your first posts, and you simply post the question and ask for help. It seems like you just want the answers.

For example, the question: Find the derivative of $f(x)=\frac{x}{1+x}$ I would treat it as a product like $f(x)=x*(x+1)^{-1}$ , and then you find the derivative by applying the product and power rules.

**VX-1** · September 24th 2008, 06:42 PM

I don't want the answers, I'm just completely clueless as to how I would approach them.

**Jhevon** · September 24th 2008, 06:58 PM

Originally Posted by VX-1

Hey guys can you help me with the questions on this Math midterm?

1.
"orthogonal" means "perpendicular." so you want all the points on
$y = 5x^3 + 5$ where the slope (that is, the derivative) is 30. please tell me you know why we need it to be 30

2.
the definition says: $f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h$

now plug in $f(x + h)$ and $f(x)$ and take the limit. (you will need to simplify before you can do that, of course)

in case you have never seen that definition, you must have used this one. it is an alternate definition equivalent to the one above.

$f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$

3.
(a) note that $\sqrt{x^2 + 1}$ increases a lot faster than $\sqrt{x}$ . alternatively, note that as $x \to \infty$ the $+1$ doesn't really matter. thus you have $\sqrt{x^2} - \srqt{x} = |x| - \sqrt{x} = x - \sqrt{x}$ for positive $x$ 's

(b) more straightforward solution: multiply by the cojugate of the denominator over itself.

alternate solution: note that $x - 4 = (\sqrt{x})^2 - 4 = (\sqrt{x} + 2)(\sqrt{x} - 2)$

4.
you want to find $a$ and $b$ so that the function is continuous and the derivative limit for both parts exist and are equal.

5.
first find $f'(x)$ (you will need the chain rule). then you can answer the problem. i hope you know how to tell something is an extreme value from the derivative, and that you know a function is increasing wherever the derivative is positive and decreasing where it is negative

6.
Let $f(x) = \frac 1{x^2 + 1}$

by tweaking our definition for derivative a bit, we obtain the linear approximation formula:

$f(x) \approx f(a) + f'(a)(x - a)$

now, $x$ is the value that we want to find $f(x)$ for, while $a$ is a value close to $x$ that we already know $f(a)$ and $f'(a)$ for

so there's how you should approach these problems. good luck!

Originally Posted by colby2152

For example, the question: Find the derivative of $f(x)=\frac{x}{1+x}$ I would treat it as a product like $f(x)=x*(x+1)^{-1}$ , and then you find the derivative by applying the product and power rules.

well, they said use the definition. i am pretty sure they are talking about using limits to calculate the derivative

**colby2152** · September 25th 2008, 05:05 AM

Originally Posted by Jhevon

well, they said use the definition. i am pretty sure they are talking about using limits to calculate the derivative

True. I missed that part of the question. You'll have to find the derivative via the tedious, tried and true method that is the definition of the derivative:

$f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

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