# Some derivative problems

I missed the lessons at school for derivatives, and I'm not really understanding things from my textbook very well, so I'm hoping someone here can help me a bit with these problems.

1. In each case, find the derivative dy/dx

a) $y = 6 - 7x$
b) $y = {x + 1}/{x - 1}$
c) $y = 3x^2$

The book didn't go into too much detail on the dy/dx thing, and so I don't really have any idea on what to do.

2. Find an equation of the straight line that is tangent to the graph of $f(x) = \sqrt{x + 1}$ and parallel to $x - 6y + 4 = 0$.

I figured that I could at least find the slope of the line, since it's parallel to $x - 6y + 4 = 0$. I calculated the slope from this to be 1/6, and since the lines are parallel, then the slope of the tangent to the graph of f(x) = \sqrt{x + 1}[/itex] must also be 1/6. I didn't really know what to do after this.

3. For each function, use the definition of the derivative to dtermine dy/dx, where a, b, c, and m are constants.

a) $y = c$
b) $y = x$
c) $y = mx + b$
c) $y = ax^2 + bx + c$

I was thinking that I might have to substitute the y in each case for the y in dy/dx, but I'm not really sure, as I said earlier, the book didn't go into dy/dx all that much.

arildno
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What's the definition of the derivative?

I don't know exactly, I know it sounds bad, but I didn't really understand the way the book explained it.

arildno
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Byrgg said:
I don't know exactly, I know it sounds bad, but I didn't really understand the way the book explained it.
well, then you should read your book again, and state PRECISELY where you have problems with the definition of the derivative.

It is meaningless to proceed before you understand that bit.

"This limit(the difference quotient I believe it is referring to) is called the derivative of f(x) at x = a."

arildno
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And what is a difference quotient?

It's the limit that's used to find the slope of a tanget.

arildno
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Is that a "difference quotient"?

That's what my textbook called it.

arildno
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No, it didn't.

Go back to the diagrams in your book.
How is the slope of a secant computed, and how is the slopes of the secants related to the slope of the tangent?

As the moving point on the line approaches the point of tangency, the slope of the secant gets closer to the slope of the tangent.

Here's what it said about the difference quotient:

${f(a + h) - f(a)}/h$

This quotient is fundamental to calculus and is referred to as the difference quotient. Therefore, the slop m of the tangent at P(a, f(a)) is lim h->0(slope of the secant PQ), which may be written as m = lim h->0 f(a + h) - f(a)/h

arildno
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Well, here it seems you have quoted the book verbatim!

Okay, let's proceed:
What's dead wrong in your notation that separates the numerator from the denominator?

What do you mean? Are you referring to the fact that I didn't have all of f(a + h) - f(a) in the numerator? If so, then it's because I don't know how to post it the proper way. It's not often that I have to use LaTex coding.

arildno
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Well, you could have done it in the standard manner: Put a parenthesis about the expression for the numerator!

Now, what is meant with the concept "limit"?

I guess the best way to describe a limit would be the number that a function approaches as the variable approaches the number specified in the limit. That was a kind of awkward and rough explanation, I'm not really sure about how to explain it very well.

arildno
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Right!

So, how can you UTILIZE the difference quotient (i.e, the expression for the slope of secants) in order to evaluate its limit as h tends to zero?

HallsofIvy
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Another definition of "derivative" you should know: "the derivative of a function f(x) at x= a is the slope of the tangent line to the graph of y= f(x) at (a, f(a))". I would be surprised if that is not in your book somewhere.

Alright, so then a derivative of a function at a point where x = a is the slope of the tangent to this function at the point (a, f(a)), is that right? I just thought I'd change the wording a bit so that it was a bit mroe clear for me.

HallsofIvy