Question

f(1) + 1^2[f(1)]^4 = 18 and f(1) = 2, differentiate

Answer 1

Original
f(x) + x^2[f(x)]^4 = 18 and f(1) = 2, find f'(1)

Answer 2

@zepdrix :)

Answer 3

\[\large\rm f(x) + x^2\cdot f(x)^4 = 18\]

Answer 4

ooh

Answer 5

So where you stuck? :U

Answer 6

now it makes a bit of sense

Answer 7

Everywhere! I think I am doing the chain part incorrectly.

Answer 8

\[y+x^2y^4=18\] implicit diff

Answer 9

Like..
= f'(x)+4[x^2(f(x))^3
but after that i'm not sure where to even go.

Answer 10

product rule and chain rule for the second part

Answer 11

Woops you forgot your product rule! :)

Answer 12

ooh. at the (x^2)[(fx)^4]?

Answer 13

now i see why \[y+x^3y^4=18\] is much nicer notation

Answer 14

\[\large\rm f+\color{royalblue}{x^2f^4}=18\]\[\large\rm f'+\color{royalblue}{(x^2)'f^4+x^2(f^4)'}=18\]Ya product rule on the second part :)

Answer 15

ahhh i keep making typos :c

Answer 16

\[\large\rm f'+\color{royalblue}{(x^2)'f^4+x^2(f^4)'}=18'\]\[\large\rm f'+\color{royalblue}{(x^2)'f^4+x^2(f^4)'}=0\]

Answer 17

\[f(x)+x^24(f(x))^3+2x(fx)^4=18\]

Answer 18

Yah maybe a little trouble with chain rule, hmm

Answer 19

\[\large\rm \frac{d}{dx}f(x)^4\quad=\quad 4f(x)^3\cdot\color{red}{f'(x)}\]

Answer 20

hmm so now what D:

Answer 21

\[\large\rm f+x^2f^4=18\]Able to get to this point ok? :)\[\large\rm f'+2xf^4+4x^2f^3f'=0\]After that, we'll umm.. oh lemme rewrite it in your notation so we can plug stuff in,\[\large\rm f'(x)+2xf(x)^4+4x^2f(x)^3f'(x)=0\]Now let's evaluate this mess at x=1.

Answer 22

Becomes something like this, ya?\[\large\rm f'(1)+2f(1)^4+4f(1)^3f'(1)=0\]

Answer 23

And they told us that \(\large\rm f(1)=2\).

Answer 24

[sorry, pc got stuck. But I am back!]

Answer 25

and now we have
f'(1)+32+32f'(1)=0
f'(1)=-32/33

Answer 26

Mmmm ya I think that's right! :)
Imma double check our work real quick.

Answer 27

yay good job! ୧ʕ•̀ᴥ•́ʔ୨

Answer 28

yayay! and we have a green check mark!
thank you so much :)