Question

help with calculus question please

Answer 1

If h(x) = f[f(x)] use the table of values for f and f ′ to find the value of h ′(1).


x f(x) f ′(x)
1 3 2
2 1 5
3 6 7

Answer 2

@mathmale

Answer 3

What could I do that would help you most at this point?

Answer 4

could you explain to me further how to get to my solution?? the problem is, that this is for a study guide like before we even learn it so i have no idea how to do it but we are still supposed to try. The best you could help me here is fully e explain to me how to get to the conclusion so i could write it down for further studies

Answer 5

Can you read me now?

Answer 6

yes i am seeing you here

Answer 7

Same here.
You have the function defined as h(x) = f( f(x) ) and you want to find the derivative of this function h(x) when x=1. Is that correct?

Answer 8

yes i believe so

Answer 9

The most important principle here is the chain rule. How comfortable are you with that rule?

Answer 10

i know it well

Answer 11

Would you mind if I gave you a sample problem for you to work on?

Answer 12

but what exactly would i be solving here with the chain rule?

Answer 13

no, if you could just help me with this specific problem that'd be most advantageous

Answer 14

please

Answer 15

Your task is to find the derivative of what we call a "composite function."

Answer 16

that's why the chain rule is needed.

Answer 17

oh ok- i haven't learned that yet

Answer 18

i know the chain rule, but not the other things. If you could show me so i could know it better that'd be great

Answer 19

Let's differentiate the composite function sin x^2 (the sine of [x^2] ).
what is the derivative? Have to apply the chain rule here. If it's not clear I'll walk you thru it.

Answer 20

ok, the first derivative would be: 2x cos(x^2)

Answer 21

Yes, that's perfect. I see you kept the argument x^2 for the cosine functi9on and then you differentiated x^2 to obtain 2x. Right.

Answer 22

yes, now what?

Answer 23

suppose i change that practice problem a bit. Instead of x^2 as argument for the sine function, I'm going to use f(x).

Find the first derivative of y=sin f(x). As you know, f(x) is just the name of some function.

Answer 24

no, can we please just stick with this one i need this one to be my example, please sir

Answer 25

i will still go through the works but please can we just stick with this

Answer 26

sure, i'm encouraging you to finish this.

y= cos f(x). Find (dy/dx).

Answer 27

ok so i am differentiating here right?

Answer 28

That's right. You're diff. cos f(x) with respect to x, using the chain rule because the input f(x) to the function cos x is itself a function.

Answer 29

would it be cos (f)?

Answer 30

Review: If y = cos x, then (dy/dx) = -sin x. Familiar?

Answer 31

oh yes thats familiar, so i wasn't right?

Answer 32

Not quite.

if y = cos x, dy/dx = - sin x.

If y = cos f(x), dy/dx = - sin f(x) * f '(x).

Again, I used the Chain rule.

Answer 33

oh ok, i am writing this down, then what?

Answer 34

You're done as far as finding the derivative of y = cos f(x). Any questions about this example? Do you see how we applied the chain rule?

Answer 35

yes i see, now can we please get back to my specific question?

Answer 36

hello?:-)

Answer 37

so sorry, I didn't see your "let's get back to my specific question."

Answer 38

h(x) is define as the function of a function, and it happens here that both functions are represented by f(x). OK so far?

Answer 39

yep, also could we please go as fast as possible its just that I have 8 more questions in which i need to be taught as well

Answer 40

Remember that the deriv. of sin x^2 is cos x^2 * 2x?

What's the derivative of h(x) = f ( f(x) )? You must first differentiate the first f(x) , keeping the same input (which is also f(x) ), and then differentiate the 2nd f(x).

Try it. Don't be afraid to make mistakies.

Answer 41

ok , it would be f^2 right?

Answer 42

I've had some troubles with my Internet connection. If there's a delay in responding, that's the reason.

How are you doing so far?

Answer 43

Once again: I have connection problems which will delay our conversation occasiopnally. Please proceed.

Answer 44

ok, so it would be f^2 right?

Answer 45

Not quite. Remember this concept: Function of another function.

y = sin f(x) represents a function (the sine) of another function ( f(x) ).
So no, there's no square in the proper ansser.

Answer 46

h(x) = f ( f(x) )

Apply the chain rule. find the derivative of the outside "f", kieeping the inside f9x) as argument, and then MULTIPLY the result by the derivative of f(x).

Answer 47

im not quite sure////...

Answer 48

h(x) = f ( f(x) )

h '(x) = f ' ( f(x) ) times f '(x). Ok with this?

Answer 49

0k

Answer 50

so we're actually done with that.

Answer 51

Now on the right side you'll see f '(x). Right? if so, find the value of f '(1) from the table.

Answer 52

ok, great, I'm not sure how to find the conclusion yet i know we are close though

Answer 53

Yes, very close. What is the value of f ' (1), from the table?

Answer 54

would it be 3?

Answer 55

I don't think so, I seem to remember it was 2. I don't have the table in front of me. Please try again. Let x=1 and determine the value of f '(1) from it.

Answer 56

oh it is 2 you are correct sir. we are so close

Answer 57

Here's where you were:

h '(x) = f ' ( f(x) ) times f '(x)

Now we'll replace that f '(x) with '2' : h '(x) = f ' ( f(x) ) times 2
OK with this?

Answer 58

If so, please find the value of f(x) when x=1.

Answer 59

i thought we agreed on 2?

Answer 60

Absolutely, we did, and 2 is correct. Notice where I put the '2' ?

Answer 61

yep

Answer 62

Wrong, I'm wrong.

f '(1) = 2, but now I'm asking you to find the value of the function f(x) itself when x=1.

Answer 63

um im not sure

Answer 64

Let me assure you that you were correct in stating that f '(1) = 2.

How did you obtain that value, 2?

Answer 65

from looking at the chart

Answer 66

If I remember right, the leftmost column of the table has the header "x" and you find the value '1' under that. Then you move to the right in that row until you're under "f '(x)" and find the value 2. does that describe what you've done?

Answer 67

yes

Answer 68

so far

Answer 69

Greate. Now do exactly the same thing for f(1) what does the table give you for f(1)?

Answer 70

its almost complete right? i need help with 8 more questions and its just taking so long, i know its my fault but most of this i haven't even learned yet

Answer 71

2?

Answer 72

I understand,b ut there's simply no way of making this go faster.

Answer 73

ok and so whats next? its almost done so ?

Answer 74

On the top of the table you see f(x) and f '(x), right? And for x=1, u see "2" underneath f '(x). What numeric value do yous ee under f(x)?

Answer 75

I've found and checked the chart and see that f(1) = 3.

Ok with that?

Answer 76

h '(x) = f ' ( f(x) ) times f '(x)

becomes

h '(x) = f ' ( f(x) ) times 2,

and this becomes h '(x) = f ' ( 3 ) times 2

All you need to do now is to determine the value of f '(x) for input '3'

Answer 77

Sorry, but without actually giving you the answer, I cannot make this any simpler or clearer.

Answer 78

f ' ( 3 ) = 7. Look in the row that begins with 3 and look under the column heading f ' (x).

Then you get h '(x) = f ' ( 3 ) times 2. Replace f '(3 ) with 7.

The answer is h '(1) = (7)(2) = 14.

Answer 79

its just 14 right? sorry I've been writing it out on paper and i have to get this thing going

Answer 80

We';ll need to hurry if you want to discuss another questin.

We could come back to the h '(1) question if it's still not clear for you.

post another question if you wish. I need to know what your intentions are.

Answer 81

ok, thanks but wait for this question the answer is just 14 correct???

Answer 82

the answer is just 14?

Answer 83

hello?

Answer 84

so you are getting delayed responses?

Answer 85

yes, that's it.

You were finding the derivative of h(x) at a certain x value (which was 1).

Answer 86

Yes, my connection quit on me again.

Answer 87

I'm willing to help with one more question. Which will it be?

Answer 88

Need to know your intentions. Want to do another problem, or have you had enough exposure to calculus for one night? ;)

Answer 89

Good night.

Answer 90

sorry, are you still here? i lost connection for a considerable time

Answer 91

IM SORRY please i STILL need help

Answer 92

Choose one of the harder of your remaining problems and post it (separately from this discussion).