Question

Let f(x)=1/x. Compute the difference quotient for f(x) at x=5 with h=0.2

Use the Power Rule to compute the derivative:
d/dt, t2/3|t=6

Answer 1

\[\large \frac{f(x+h)-f(x)}{h} = \frac{\frac{1}{x+h}-\frac{1}{x}}{h}\]

Now you just need to simply substitute \(x=5\) and \(h=0.2\) into the above which should be straightforward (I hope)!

Answer 2

okay, so 1/5.2 - 1/5 = .2??

Answer 3

so would the derivative be 1?

Answer 4

Well more like 1/5.2 - 1/5 , and then divide that whole result by 0.2

Answer 5

hm that subtraction shouldn't give 0.2 o_O

Answer 6

what are you getting?

Answer 7

\(\dfrac{1}{5.2}-\dfrac{1}{5}=-0.007692\)

Answer 8

yeah that's what i got on my calculator, so i do -.007692/.2?

Answer 9

yep \(\checkmark\)

Answer 10

i'm getting -.384615385

Answer 11

-.0384615385**

Answer 12

yup !

Answer 13

how about the second part of the problem?

Answer 14

Use the Power Rule to compute the derivative:
d/dt, t2/3|t=6

Answer 15

I just want to be clear since the notation doesn't translate well in regular text... the problem is asking for:
\[ \frac{d}{dt}\, \left.\frac{t^2}{3}\right|_{t=6} \] ?

Answer 16

that's correct, except it should be t^(2/3)

Answer 17

ohh ok I see

Answer 18

So recall that the power rule states:
\[\frac{d}{dt} t^n=nt^{n-1} \], here your \(n=\dfrac{2}{3}\)... so:

\[\large \frac{d}{dt} t^{\frac{2}{3}}=\frac{2}{3}t^{\frac{2}{3}-1}=\frac{2}{3}t^{-\frac{1}{3}}\]

Answer 19

Now, since they want the derivative evaluated at t=6, just substitute t=6 into the above

Answer 20

so 0.366881?

Answer 21

ya that would be good :) (But if your question requires an exact answer rather than a decimal approximation... then you wouldn't need to evaluate it further)

Answer 22

thank u!

Answer 23

your welcome =]