Question

f(x) + x^2[f(x)]^3 = 10
f(1)=2
find f'(1)

Answer 1

please help!

Answer 2

\(\Large\bf \color{royalblue}{\text{Welcome to OpenStudy! :)}}\)

Answer 3

So derivative huh? Where ya stuck?

Answer 4

I have f'(x)+2x[f(x)]^3 +x^2 (3[f(x)]^2 * f'(x))

Answer 5

Ok very good.
And the derivative of the `right side` ?

Answer 6

0

Answer 7

\[\Large\bf\sf f(x)+x^2\left[f(x)\right]^3\quad=\quad 10\]
\[\Large\bf\sf f'(x)+2x\left[f(x)\right]^3+3x^2\left[f(x)\right]^2\;f'(x)\quad=\quad 0\]Ok good, looks like you've got a great so far.

Answer 8

Now let's evaluate the expression at x=1.

Answer 9

but f'(1) is unknown

Answer 10

\[\Large\bf\sf f'(\color{orangered}{1})+2(\color{orangered}{1})\left[f(\color{orangered}{1})\right]^3+3(\color{orangered}{1})^2\left[f(\color{orangered}{1})\right]^2\;f'(\color{orangered}{1})\quad=\quad 0\]

Answer 11

Correct. After we everything `else` plugged in, we're going to try and `solve for` f'(1).
We want to isolate the f'(1).

Answer 12

how do you isolate f'(1)?

Answer 13

We'll have to take advantage of factoring and some other simple algebra.
Let's not worry about that just yet.
Let's get all the coefficients simplified first.

Answer 14

We need to use the initial data that they gave us,\[\Large\bf\sf f(1)=2\]

Answer 15

yes I understand that much and after simplifying.

f'(1)+16+12*f'(1)=0

Answer 16

is that correct?

Answer 17

Mmm good good.

Answer 18

now I am really stuck

Answer 19

\[\Large\bf\sf \color{royalblue}{f'(1)}+16+\color{royalblue}{12\;f'(1)}=0\]Combine like-terms.

Answer 20

If f'(1) is an apple.

You currently have one apple + 16 + 12apples = 0

Answer 21

so 13 apples?

Answer 22

\[\Large\bf\sf 16+13\;f'(1)=0\]Ok good.

Answer 23

How do we solve for f'(1) from here? :o

Answer 24

so 13/16?

Answer 25

Hmm I think your fraction is upside down +_+ and missing a negative, yes?

Answer 26

Careful with that quick mental math lol

Answer 27

lol yeah I should know better than to do mental math

Answer 28

-1.23?

Answer 29

f'(1) = -16/13

Yessss good job \c:/

Answer 30

thank you can you possibly help me with more?? :)

Answer 31

Close this thread down.
Open a new one so we have a nice clean space to work with :O
Also, that way in case I get too busy, someone else can see your question at the top of the list :D

Answer 32

don't help anyone else! Just me please! :)

Answer 33

Lol there are plenty of smart people who can help you on here XD
Don't worry.

But if you're waiting long, try typing @zepdrix or @ followed by someone else in the lobby with a 99 by their name.

It sends an alert to them so they can quickly get to your question.