Question

I need a little help with finding the derivative. if you could please explain step by step. I would really appreciate it.
Find the derivative of f(x) = 5/x at x = -1

Answer 1

i am only in pre-calc btw

Answer 2

just in case you use a calc shortcut

Answer 3

not sure ... how to cover it then.... because using Limits, would be calc, and using the power rule will also be calc.....

Answer 4

hmmm well.. actually limits would be precalc so

Answer 5

what was included in the lesson was (f(x+h)-f(x))/h

Answer 6

to find derivative

Answer 7

\(\bf lim_{x\to -1}\quad \cfrac{5}{x}\) will do IMO

Answer 8

there are answer choices though

Answer 9

and thats not the derivative

Answer 10

im just not sure how to move everything out of the numerator's denominator

Answer 11

have you been taught how to do \[f'(a) = \lim_{x \rightarrow a} \frac{ f(x)- f(a) }{ x-a }\]

Answer 12

no

Answer 13

\(\bf lim_{x\to -1}\quad \cfrac{5}{x}\implies lim_{h\to 0}\quad \cfrac{f(x+h)-f(x)}{h}\\ \quad \\
\implies lim_{h\to 0}\quad \cfrac{\frac{5}{x+h}-\frac{5}{x}}{h}\implies lim_{h\to 0}\quad \cfrac{\frac{5x-5x-5h}{x(x+h)}}{h}\\ \quad \\
lim_{h\to 0}\quad \cfrac{\frac{-5h}{x(x+h)}}{h}\implies lim_{h\to 0}\quad \cfrac{-5h}{x(x+h)}\cdot \cfrac{1}{h}\implies \cfrac{-5}{x(x+h)}\)

Answer 14

and as you can see, once h = 0, then we end up with \(\bf \implies \cfrac{-5}{x(x+0)}\implies \cfrac{-5}{x^2}\)

Answer 15

i thought that you cant set h to 0 when its in the denominator

Answer 16

well, yes, you're correct IF it makes the rational undefined, after the factoring and simplifying it does not