Question

Suppose that
F = f composition g,

H(x) = x(x − 2f(x))^4,

g(1) = −2,

g '(1) = 4,

f(1) = 1,

f '(1) = 5,
and
f '(−2) = −1.

Find F '(1) and H'(1).

Answer 1

Use the chainrule to find F'

Use the sumrule, productrule and the chainrule to find H'

substitue the given values into these and your're done

Answer 2

@Let.Van

Answer 3

I’m just learning this right now, but here is my attempt.

Since \[F'(x) = f(g(x))\] I think it would be clearer to see what F'(x) is by applying the Chain Rule:

\[\frac{ d }{ dx }F(x) = \frac{ d }{ dg }f(g(x)) \frac{ d }{ dx }g(x)\]

Thus you need to figure out what the derivatives of f(g(x)) and g(x) are when x=1, in order to find F’(1).

To find H’(x) I’m not completely sure, but since you are going to be dealing with x=1 all the time, and the only function inside H(x) is f(x), probably you’ll need to find the values of f(1) and f’(1). For the sake of simplicity, I would replace f(x) by f and f’(x) by f’ when needed.

I would expand the binomial to the 4th power, and then take the derivatives. The coefficients will be just constants, therefore can be place outside the derivate of xf that is needed to find. As an example, it'll be necessary to take the derivative of (x^2)(f^3), and so the product rule must be applied. But it also can be noticed that f is a function, so the exponent rule (for derivatives) must be used to differentiate f^3 as well.

I hope not to have made a mistake.

Answer 4

Ups, it was F(x). Sorry about that.

I found H(x) somewhat more difficult to solve because I considered expanding the binomial first. It seems not possible to me to distribute the x among what is inside the parenthesis, because of the exponent. And so after applying the binomial theorem, and then distributing that x, I got:

\[H(x) = x^5 -8x^4f + 24x^3f^2 – 32 x^2f^3 + 16xf^4\]
(I used f instead of f(x) in order to have a much simpler expression)

The following step is to take the derivatives of the variables xf (raised to some power) of each term. But since f is not just a variable but a function, it’ll be necessary to apply the exponent rule too:

\[\frac{ d }{ dx }[f(x)]^n = nf(x)^{n-1} \frac{ d }{ dx }f\]

After doing so I get:

\[H’(x) = -35x^4 + 208x^3 – 408x^2 + 256x + 16\]
\[H’(1) = -35 + 208 – 408 + 256 + 16\]
\[H’(1) = 37\]

Answer 5

@abtster

Answer 6

Yes, sorry about that, I got the parenthesis wrong, you were right

Answer 7

But another approach is also possible

H(x) = x(x - 2f)⁴

Multiplication rule (hk) = h'k + hk' =
H'= (x - 2f)⁴ + x[(x - 2f)⁴]'

Chainrule
[(x - 2f)⁴]' = 4(x - 2f)³[x - 2f]'

[x - 2f]' = 1 - 2f'
[(x - 2f)⁴]' = 4(x - 2f)³(1 - 2f')

H'= (x - 2f)⁴ + 4x(x - 2f)³(1 - 2f')

Now substite
f(1)=1
f'(1) = 5
x=1

H'(1) = (1 - 2)⁴ + 4(1 - 2)³(1 - 2*5) = 1 + 4(-1)³(-9) = 37

Answer 8

Thankyou, for all the effort you two have put into explaining this problem. I manage to figure out how to solve for H'(1)
However, part one F = f(g(x)) still bothers me, I'm not sure how to approach it as g(x) is not define(?)

Answer 9

@abtster
@Joq

Answer 10

Well f(x) is not defined either
it is from the the given values that you can find values at the specific value of x for which the relevant values are given
Joq gave the Chain Rule for derivation, that is for derivation of composit functions
\[\frac{ df(g(x)) }{ dx} = \frac{ df(g(x)) }{ dg(x) }\frac{ dg(x) }{ dx }\]

so you differentiate as if g(x) is the vvariable like x, but you have to compensate for that by multiplying by the derivative of g(x)

I find it usually quite easy to apply

Answer 11

if you find the g(x) confusing leave out the x at first
\[\frac{ df(g(x)) }{ dx} = \frac{ df(g)) }{ dg}\frac{ dg(x) }{ dx }\]

Answer 12

\[\frac{ df(g(x)) }{ dx} = \frac{ df(g) }{ dg }\frac{ dg(x) }{ dx }\]

forgot to wipe out one of the parenthesis