Finding slope of a function: y = 1/x at x = 2

[Euclid Elements]

Hello, I've been struggling with a function slope problem lately. The question is:

\(\displaystyle \mbox{What is the slope of the function }\, y\, =\, \dfrac{1}{x}\, \mbox{ at }\, x\, =\, 2\, ?\)

I got to the point where I had to perform the operation 1 / (x+dt) - 1/x, but here I got stuck. Any help would be appreciated. Thanks!

 

[Deleted member 4993]

Hello, I've been struggling with a problem lately. The question is:

I got to the point where I had to perform the operation 1 / (x+dt) - 1/x, but here I got stuck. Any help would be appreciated. Thanks!

Do you know:

\(\displaystyle \dfrac{d}{dx}(x^n) = n * x^{n-1}\)

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

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[Ishuda]

Hello, I've been struggling with a problem lately. The question is:

I got to the point where I had to perform the operation 1 / (x+dt) - 1/x, but here I got stuck. Any help would be appreciated. Thanks!

\(\displaystyle \frac{f(x+dx)-f(x)}{dx}=\frac{\frac{1}{x+dx}-\frac{1}{x}}{dx}\)

=\(\displaystyle \frac{\frac{x}{x(x+dx)}-\frac{(x+dx)}{x(x+dx)}}{dx}\)

=\(\displaystyle \frac{\frac{x-(x+dx)}{x(x+dx)}}{dx}\)

=\(\displaystyle \frac{\frac{-dx}{x(x+dx)}}{dx}\)
= ...
Can you continue from there?

 

[stapel]

I got to the point where I had to perform the operation 1 / (x+dt) - 1/x, but here I got stuck.

Would it be correct to assume, from the form of the expression you've posted, that you're supposed to "find the derivative from the limit definition" (and that you're taking the limit as dx goes to zero)? If so, then:

Hello, I've been struggling with a function slope problem lately. The question is:

\(\displaystyle \mbox{What is the slope of the function }\, y\, =\, \dfrac{1}{x}\, \mbox{ at }\, x\, =\, 2\, ?\)

You're trying to do the following (if I've guessed your meaning correctly):

. . . . .\(\displaystyle \displaystyle y'\, =\, f'(x)\, =\, \lim_{dx\, \rightarrow \,0}\, \dfrac{f(x\, +\, dx)\, -\, f(x)}{dx}\, =\, \dfrac{\left(\, \dfrac{1}{x\, +\, dx}\, -\, \dfrac{1}{x}\, \right)}{dx}\)

If this is correct, then try working with the complex-fraction expression. (You can return to the limit stuff later.) A good first step is to convert to common denominators. In this case:

. . . . .\(\displaystyle \dfrac{\left(\, \dfrac{x}{x\, (x\, +\, dx)}\, -\, \dfrac{x\, +\, dx}{x\, (x\, +\, dx)}\, \right)}{dx}\, =\, \dfrac{\left(\, \dfrac{x\, -\, x\, -\, dx}{x\, (x\, +\, dx)}\, \right)}{dx}\, =\, \dfrac{\left(\, \dfrac{-dx}{x\, (x\, +\, dx)}\, \right)}{dx}\)

Where does this lead you?