Question

Find the derivative with respect to x of the given combination at the given value of x

g(x+ f(x)), x = 3

Answer 1

HOw is the answer -28?

Answer 2

g(3+ 1) = g(4)

Answer 3

Then I'm not sure what to do from there.

Answer 4

first you need to differentiate...then put x=3, not before.

Answer 5

whats derivative of g(x+f(x)) ??

Answer 6

-4?

Answer 7

using chain rule,
\([g(x+f(x))]'=g'(x+f(x))[(x+f(x))]'\)
got this ?

Answer 8

Let me try to work it out using that formula... I actually don't remember seeing this before.

Answer 9

chain rule is :
\([g(f(x))]'= g'(f(x))[f'(x)]\)

Answer 10

Still a little confused...

[g(x+f(x))]′=g′(x+f(x))[(x+f(x))]′
= g'(4) * 4 ?

Answer 11

lets go term by term
you put x=3,
f(3) =....?
x+ f(x) = 3+f(3)=... ?

Answer 12

i think u did that...and comes out to be 4

Answer 13

g'(4) from the table =... ?

Answer 14

g'(4) = -4

Answer 15

yes, thats correct ,
now [x+f(x)]' = 1+ f'(x) = 1+ f'(3)=...?

Answer 16

7

Answer 17

how u got , [(x+f(x))]′ at x=3 as 4 ??

Answer 18

you got how thats 7 ??

Answer 19

now [x+f(x)]' = 1+ f'(x) = 1+ f'(3)=...?
1+ 6 = 7 ?
f'(3) = 6

Answer 20

yeah, i meant to ask, did you get how [x+f(x)]' = 1+ f'(x) ?

Answer 21

Yeah, from the formula...
The second part is [(x+f(x))]′ so we find that and multiply it by g′(x+f(x))

[g(x+f(x))]′=g′(x+f(x)) [(x+f(x))]′

Answer 22

oh but why did x turn to 1 ?

Answer 23

i didn't turn, i differentiated,
\(\dfrac{d}{dx}(x)=1\)

Answer 24

[x+f(x)]' = x'+f'(x) = 1+ f'(x)

Answer 25

as i said, we first differentiate, then put x=..
any more doubts ?

Answer 26

Ah okay. I get it now. You been very helpful! Thank you so much!

Answer 27

welcome so much! :)