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anonymous
 one year ago
Precalculus help!!
1. Find the derivative of f(x) = 6 divided by x at x = 2.
and
2. Find the derivative of f(x) = 4x + 7 at x = 5.
Thank you. Need help solving, will medal.
anonymous
 one year ago
Precalculus help!! 1. Find the derivative of f(x) = 6 divided by x at x = 2. and 2. Find the derivative of f(x) = 4x + 7 at x = 5. Thank you. Need help solving, will medal.

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freckles
 one year ago
Best ResponseYou've already chosen the best response.2I bet you are practicing using the formal definition and not the short cuts right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you could use either one of these: \[f'(a)=\lim_{h \rightarrow 0}\frac{f(a+h)f(a)}{h} \text{ or } f'(a)=\lim_{x \rightarrow a} \frac{f(x)f(a)}{xa}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2they are the same definition just written a little different

freckles
 one year ago
Best ResponseYou've already chosen the best response.2which one are you more comfortable with?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure, I'm still confused with both.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Ok we could do question one together both ways and you can decide which way you want to use for the second and I will check your work for that one

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[a=2 \\ f(x)=\frac{6}{x} \\ f(a)=f(2)=\frac{6}{2}=3 \\ f(a+h)=f(2+h)=\frac{6}{2+h}\] first do you understand how I got f(a+h) and f(a) for the first one?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Ok I'm going to input them into the formula

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f'(a)=f'(2)=\lim_{h \rightarrow 0}\frac{\frac{6}{2+h}(3)}{h}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f'(2)=\lim_{h \rightarrow 0}\frac{\frac{6}{2+h}+3}{h}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2ok the trick is to get that h on the bottom to cancel somehow so we can actually plug in h=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.2notice we have a compound fraction

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that is a fraction inside a bigger fraction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hold on super quick, writing this down.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2we want to see if we can "simplify" the fraction that means get rid of the compound fraction notice the little fraction has denominator (2+h) multiply top and bottom by (2+h) this will make our compound fraction into a "simple" fraction do you want to try multiplying top and bottom by (2+h)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and let me know what the result is just doing that one step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do I multiply the top and bottom of just the little fraction?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2multiply top and bottom by (2+h) of the big fraction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would it be 9/h(2+h) ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that isn't a bad try ... I think the 9 is just a bit off...should be 3 there \[\lim_{h \rightarrow 0} \frac{\frac{6}{2+h}+3}{h} \cdot \frac{2+h}{2+h}=?\] \[\lim_{h \rightarrow 0} \frac{\frac{6(2+h)}{2+h}+3(2+h)}{h(2+h)} \\ =\lim_{h \rightarrow 0}\frac{6+3(2+h)}{h(2+h)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\lim_{h \rightarrow 0} \frac{66+3h}{h(2+h)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok, I see. So it would be 3h/h(2+h)?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\lim_{h \rightarrow 0}\frac{3h}{h(2+h)}\] you should recall h/h=1 (we can say this since h is not 0)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\lim_{h \rightarrow 0}\frac{3 \cancel{h}}{\cancel{h}(2+h)}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So then we are left with 3/2+h

freckles
 one year ago
Best ResponseYou've already chosen the best response.2we are going to have no problems pluggin in 0 for h now because this function 3/(2+h) actually exists at h=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so what do you think f'(2)=?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But how do you know that the function exists at 0?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2because the bottom is not 0 when h is 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok got it, so would f'(2) = 3/2?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2doesn't matter where you put the negative at

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Could you check my other question once I solve it?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2k.. while you are working it I'm going to do the first one again but use the other formula I wrote

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is the second one f(5) = 4?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[f'(a)=\lim_{x \rightarrow a}\frac{f(x)f(a)}{xa} \\ a=2 \\ f(x)=\frac{6}{x} \\ f(a)=f(2)=\frac{6}{2}=3 \\ f'(a)=f'(2)=\lim_{x \rightarrow 2} \frac{\frac{6}{x}+3}{x+2} \\ \text{ multiply \top and bottom by } x \\ f'(2)=\lim_{x \rightarrow 2} \frac{6+3(x)}{(x+2)(x)} \\ f'(2)=\lim_{x \rightarrow a}\frac{3(x+2)}{(x+2)(x)} \\ \text{ notice} \frac{x+2}{x+2}=1 \text{ for when } x \neq 2 \\ f'(2)=\lim_{x \rightarrow 2} \frac{3}{x} =\frac{3}{2} \text{ or } \frac{3}{2} \text{ or } \frac{3}{2} \text{ or } 1.5\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2oops one sec let me check

freckles
 one year ago
Best ResponseYou've already chosen the best response.2"2. Find the derivative of f(x) = 4x + 7 at x = 5." f(5)=4(5)+7=20+7=27

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok, I'm used to the first version we used though. But the one you just used also makes a lot of sense. (:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, and for f(a+h) I got it = to 27+h

freckles
 one year ago
Best ResponseYou've already chosen the best response.2did you mean f'(5)=4?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes that's what I meant.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Could you check this one too? The position of an object at time t is given by s(t) = 8  9t. Find the instantaneous velocity at t = 1 by finding the derivative.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yeah this question is basically the same as the last questions you asked just asked it in word problem form

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would it be: f'(1) = 9?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the velocity=derivative of position v(t)=s'(t) They are using to find s'(1)

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you are missing a sign

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[v(1)=s'(1)=\lim_{h \rightarrow 0}\frac{s(1+h)s(1)}{h}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My bad, didn't catch the negative sign.

freckles
 one year ago
Best ResponseYou've already chosen the best response.2are you using short cuts? or the long way?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2because you found that really fast

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The long way but using mental math

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have the steps written down though, the long way. Would this be ok for an answer: First, I used: s(t) = 8  9t and t=1 to solve for (a+h) and (a). My 'a' term would be 1 in this case. f(a) = 8  9t = 8  9(1) = 17 f(a+h) = 8  9t = 8  9(1+h) = 17  9h Then I used the equation: f'(a) = lim[f(a+h)  f(a)/h] = f'(a) = 17  9h  (17)/h = 9h/h = 9 Thus I found the derivative to be: f'(1) = 9

freckles
 one year ago
Best ResponseYou've already chosen the best response.2s(t) = 8  9t notice the slope of s(t) is 9 the derivative of s(t) is 9

freckles
 one year ago
Best ResponseYou've already chosen the best response.2but yeah your teacher probably wants you to use that formal definition of derivative thing but you know we remember from algebra that the slope of f(x)=mx+b is m and we learn from calculus that the slope is the derivative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, but is the way I showed my work using the formal definition?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2only thing I would say and it is really not apart of your work "s(t) = 8  9t and t=1 to solve for (a+h) and (a). My 'a' term would be 1 in this case. :" I would say f(a+h) and f(a) but I knew what you meant when you said this

freckles
 one year ago
Best ResponseYou've already chosen the best response.2oops s(a+h) and s(a) since our function name is s not f

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you could change all those f letters to s letters actually

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for all your help!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I will be doing another test similar if you could help me later on? @freckles Maybe in an hour or so.
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