Question

Use the following table to find the value of h '(2) if h(x) = f[g(x)]:

Answer 1

1st thing is to figure out h'(x) by using chain rule: https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/chainruledirectory/ChainRule.html

Answer 2

(just for reference... will see if I am capable of helping)

Answer 3

You want to find \(\color{#000000 }{ \displaystyle h'(2) }\)
(the derivative of h(x), evaluated at x=2).

You are given that \(\color{#000000 }{ \displaystyle h(x)=f[g(x)] }\)

Answer 4

Can you find the derivative of \(\color{#000000 }{ \displaystyle f[g(x)] }\) (using the Chain Rule)?

Answer 5

is it 0?

Answer 6

I don't know how you came up with that result, because even if you plugged the equivalent values you still wouldn't have got that.

Answer 7

Let's perhaps start from this... (sorry that I have to do this)

do you know the Chain Rule?

Answer 8

im lost can you explain it to me?

Answer 9

The chain rule or the problem?

Answer 10

or both?

Answer 11

both

Answer 12

Well, I will start from explaining the chain rule, from the examples and from the "what" ...

you can prove the Chain Rule multiple ways, but I don't think we need to worry about proving here....

Answer 13

For example, you have the following function:
\(\large\color{#000000 }{ \displaystyle f(x)=\ln\left(x^2+4\right) }\)

And say you want to differentiate this function,
then you need to know two things:

[1] The derivative of ln(x) is 1/x.
[2] The derivative of x²+4 is 2x.

So, as you differentiate you differentate the function
as if is just ln(x) (without the inner function, x²+4)

And then, multiply times the derivative of this
inner function (and the inner function, once again is, x²+4)

So, the derivative f'(x) goes as follows:

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{1}{x^2+4}\times 2x }\)

which simplifies to

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{2x}{x^2+4}}\)

Answer 14

I will give you another example of the chain rule, and let me know if you understand the Chain Rule at least somewhat, or if you want more examples more theory, proves or other information about the chain rule....

Answer 15

For example, you have the following function:
\(\large\color{#000000 }{ \displaystyle f(x)=\sqrt{2x^3+12x-3} }\)

You want to differentiate this function, and
to do this, you will need to know two things:


[1] The derivative of \(\sqrt{x}\) is \(\small\dfrac{1}{2\sqrt{x}}\).\({\tiny\\[1.2em]}\)
[2] The derivative of 2x³+12x-3 is 6x²+12.
\({\tiny\\[1.2em]}\)
So, as you differentiate you differentate the function
as if is just ln(x) (without the inner function, x²+4)

And then you again you will differentiate the
functionas if it is just √x.

The derivative of f(x) will be:

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{1}{2\sqrt{2x^3+12x-3}}\times(6x^2+12) }\)

which simplifies to

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{6x^2+12}{2\sqrt{2x^3+12x-3}} }\)

\(\large\color{#000000 }{ \displaystyle f'(x)=\frac{3x^2+6}{\sqrt{2x^3+12x-3}} }\)

Answer 16

ok that makes sense now I'm trying to figure out how that applies to the chart

Answer 17

ok, I will tell you in a minute :)

Answer 18

So, suppose that you have two functions:

A differentiable function \(\large\color{#000000 }{ \displaystyle r(x) }\)

And a composite function T, such that:
\(\large\color{#000000 }{ \displaystyle {\rm T}\left[~r(x)~\right] }\)

Answer 19

The derivative or r(x), would just be r'(x).
And the derivative of T(x) would also be just T'(x).


HOWEVER, when you are taking T\([\color{blue}{r(x) }]\),

then the derivative would be:
T\('[\color{blue}{r(x) }]\times r'(x)\)

Answer 20

This is via the Chain Rule principle,
i.e. you differentiates the T(r(x)) "as is it is T(x)"
(saying, that you treat r(x) as just a variable x, at first)

And then, you multiply times the derivatiev of the inner function
(or times the derivative of r(x) - which is r'(x))

Answer 21

differentiated**
(excuse my grammar right now)

Answer 22

By the same tocken, what would be the derivative of \(f[\color{blue}{g(x) }]\) ?

Answer 23

so g' (x)?

Answer 24

The derivative of \(f[\color{blue}{g(x) }]\) would be, \(f'[\color{blue}{g(x) }]\times \color{blue}{g'(x) }\)

Answer 25

let's review the chain rule again:

You know that:
the derivative of \(f[\color{blue}{x}]\) would be, \(f'[\color{blue}{x }]\)
the derivative of \(g(\color{blue}{x})\) would be, \(g'(\color{blue}{x })\)

THEREFORE,
the derivative of \(f[\color{blue}{g(x)}]\) would be ...?

(you may refer to previous examples)

Answer 26

\(\large\color{black}{ \large{ \begin{array}{| l | c | r |}
\hline \bf~x~ & \bf a & \bf b & \bf c \\
\hline
\scr~f(x)~ & \scr \ell_1 & \scr \ell_2& \scr \ell_3 \\
\hline
\scr~g(x)~ & \scr c & a~& \scr b~ \\
\hline
\scr~f'(x)~ & \scr \ell_4& \scr \ell_5& \scr \ell_6 \\
\hline
\scr~g'(x) ~ & \scr \ell_7 & \scr \ell_8& \ell_9~ \\
\hline \end{array} } }\)

Where the \(\color{black}{\ell_{\rm number}}\)
denote arbitrary constants that are real numbers.

(Want to be absract and yet explanatory,
hope it isn't too hard to read)

Now, you are given that,
\(\large\color{#000000 }{ \displaystyle {\rm Z}{(x)=f\left[{\tiny~}g(x){\tiny~}\right]} }\)

And suppose you want to find \(\large\color{#000000 }{ \displaystyle {\rm Z}{{\tiny~}'(a)} }\).

You will use the chain rule principal
to find the derivative Z'(x). But, you
know that:
\(\large\color{#000000 }{ \displaystyle {\rm Z}{(x)=f\left[{\tiny~}g(x){\tiny~}\right]} }\)

So, really you are differentiating the
\(\large\color{#000000 }{ \displaystyle f\left[{\tiny~}g(x){\tiny~}\right] }\)

And you know that:
[1] the derivative of any function \(\large\color{#000000 }{ \displaystyle f\left(x\right) }\) is \(\large\color{#000000 }{ \displaystyle f{\tiny~}'\left(x\right) }\)

[2 the derivative of inner function \(\large\color{#000000 }{ \displaystyle g\left(x\right) }\) is \(\large\color{#000000 }{ \displaystyle g{\tiny~}'\left(x\right) }\)

THEREFORE, the derivative of \(\large\color{#000000 }{ \displaystyle f\left[{\tiny~}g(x){\tiny~}\right] }\)
is, \(\large\color{#000000 }{ \displaystyle f{\tiny~}'\left[{\tiny~}g(x){\tiny~}\right] \times g{\tiny~}'(x) }\)

So, you know that
\(\large\color{#000000 }{ \displaystyle {\rm Z}{\tiny~}'(x)=f{\tiny~}'\left[{\tiny~}g(x){\tiny~}\right] \times g{\tiny~}'(x) }\)

And therefore, you know that
\(\large\color{#000000 }{ \displaystyle {\rm Z}{\tiny~}'(a)=f{\tiny~}'\left[{\tiny~}g(a){\tiny~}\right] \times g{\tiny~}'(a) }\)

You are given that g(a)=c, so you have,
\(\large\color{#000000 }{ \displaystyle {\rm Z}{\tiny~}'(a)=f{\tiny~}'(c) \times g{\tiny~}'(a) }\)

Then, lastly, you are given that:
\(f'(c)=\ell_6\) and \(g'(a)=\ell_7\)

So your final answer is,
\(\large\color{#000000 }{ \displaystyle {\rm Z}{\tiny~}'(a)=\ell_6\times \ell_7 }\)

Answer 27

that is just an example.

Answer 28

but a general one...
this should clarify everything.

good luck