Question

where do i begin with this?

Answer 1

f(x)=sqrt(x^2+25) + sqrt((x-3)^2+16)
find f'(3)

Answer 2

chain rule?

Answer 3

\[f(x)=\sqrt{x^2+25}+\sqrt{(x-3)^2+16}\]??

Answer 4

\[f(x)=\sqrt{x^2+25}+\sqrt{x^2-6x+25}\]

Answer 5

\[f'(x)=\frac{2x}{2\sqrt{x^2+25}}+\frac{2x-6}{2\sqrt{x^2-6x+25}}\]

Answer 6

cancel the 2, replace x by 3

Answer 7

how did u get to that last step

Answer 8

He used the product rule combined with the chain rule. Notice that \(\sqrt{x} = x^{1/2}\)

Answer 9

final answer is 3/sqrt(34) + -3/4 ???

Answer 10

@satellite73 ? @KingGeorge ?

Answer 11

@satellite73

Answer 12

sam can u answer my question? satellite seems to be busy or something.

Answer 13

Differentiate the two trinomial

Answer 14

so differentiate x^2 - 6x + 25 before plugging in 3?

Answer 15

differentiate f(x)=sqrt(x^2+25) + sqrt((x-3)^2+16)
then, use f(3) to x and get the answer

Answer 16

so 2x and 2x-6???

Answer 17

then 2(3) and 2(3)-6

Answer 18

yes, when
f(3), it is saying that substitute 3 into x

Answer 19

i get 3/sqrt(6) + -3/sqrt(0)

Answer 20

doesnt look right

Answer 21

@amistre64 , @satellite73 ...i need a final answer here too

Answer 22

@Mertsj
your help would be appreciated too

Answer 23

Do you agree with Satellite's first derivative?

Answer 24

i dont know what to believe lol

Answer 25

well if you can't believe Satellite, you can't believe anybody.
Anyway if you replace x with 3 in the first derivative, you get 3sqrt34/34

Answer 26

\[f(x)=(x^2+25)^{1/2} + ((x-3)^2+16)^{1/2}\]
\[f'(x)=\frac12(x^2+25)^{-1/2}*(x^2+25)' + \frac12((x-3)^2+16)^{1/2}*((x-3)^2+16)'\]
\[f'(x)=\frac12(x^2+25)^{-1/2}*(2x) + \frac12((x-3)^2+16)^{1/2}*(2(x-3))\]

Answer 27

then plug in 3?

Answer 28

\[f'(3)=\frac12(3^2+25)^{-1/2}*(2.3) + \frac12((3-3)^2+16)^{1/2}*(2(3-3))\]
\[f'(3)=\frac12(9+25)^{-1/2}*(6) + \frac12((0)^2+16)^{1/2}*(2(0))\]
\[f'(3)=\frac12(9+25)^{-1/2}*(6) + \bcancel{\cancel{\frac12((0)^2+16)^{1/2}*(2(0))}}^{\ 0}\]
\[f'(3)=3(34)^{-1/2}\]

Answer 29

so yeah\[\frac{3\sqrt{34}}{34}\]