here is the question and my answers, but what is the second part ??

Question

Q: Differentiate with respect to x:

a) g(x) = 1 -1/x

= 1/x^2 ?

b) h(x) = 2e^x

= 2e^x

c) j(x) = sin(x) + cos(x)

= cos(x) - sin(x) ?

d) k(x) = ln (x)

= 1/x ?

THE SECOND PART OF THE QUESTION:

says to graph the functions above (which I have done), and compare the rate of change for each of the functions when x is close to 1 ???

Answer 1

not sure

Answer 2

You have done a lot of work on this already, so using your results:-

a) g `(x) = 1 - 1/x where g `(x) is the rate of change of the function g(x) at x.
rate of change at x = 1 is given by:-
g ` (1) = 1 - 1/1 = 0

b) similarly h `(x) = 2e^x and h `(1) = 2 e

c) j `(x) = cos x - sin x and j`(1) = cos1 - sin1 where cos1 is cos1radian and sin1 = sin1 radian

d) k`(x) = 1/x and as x approaches 1, k`(x) approaches 0

Hope that this contribution is of use to you in completing the question.

Answer 3

The rate of change of each function is given by the derivative, which you've worked out. So just evaluate the derivates at x = 1. The numbers you get shouldmatch what's happening on the graph (i.e. the slope of the tangent line at x = 1).

Answer 4

I think for the second part of the question, what you're being asked to solve for is this:

lim [f(x) - f(1) ] / (x - 1)
x -> 1

{Or, some other function in place of f(x).}

a) g(x) = 1 - 1/x
g'(1) =
lim [(g(x) - g(1)] / (x - 1)
x -> 1

lim [(1 - 1/x) - (1 - 1/1)] / (x - 1)
x -> 1

lim [(1 - 1/x) - (1 - 1)] / (x - 1)
x -> 1

lim [(1 - 1/x)] / (x - 1)
x -> 1

Multiplying top and bottom by x,

lim [(x - 1) / {x(x - 1)}]
x -> 1

Cancelling the (x - 1) on the top and bottom, we get

lim (1/x)
x -> 1

Which is just equal to 1/1, or 1.

Given that you've shown effort in doing these questions, I'm 100% confident you can figure out the rest.

Answer 5

i think you just gotta say increasing / decreasing / no change that kinda thing. like for the last one it kinda approaches a vertical asymptote cos as x approaches 1 the derivative approaches infinity right so just explain that?

oh wait.

your derivative IS your rate of change .. so just put 1 in the equation and if its positive it means theres positive rate of change (increase) if its negative its decreasing etc etc ..

should be that i think ..

Answer 6

yeah, that question sucks. if it did not require so much thought, i woul;d explain it

Good luck!!