Question

How to get the derivative of this:

Answer 1

\[ f(x) =\frac{ 1 }{ 4 }\sqrt{(3^{2}+x ^{2}} +\frac{ 1 }{ 6 }(12-x)\]

Answer 2

This is what I'm trying to solve.

Answer 3

Im looking for the minimum so i built this function and I think I need to solve for the zeros but im not sure how. Does anyone have any ideas?

Answer 4

Get the derivative and set it = 0

Answer 5

Yes. I cant get the derivative. :(

Answer 6

To do the derivative you should be applying the chain rule for the first part of the expression and then it's just a sum.

Answer 7

Ohh I didn't see that. Thanks!

Answer 8

It'll be easier if you set it as \[f(x) = \frac{ 1 }{ 4 }(3x^2+x^2)^{1/2}+\frac{ 1 }{ 6 }(12-x)\]

Answer 9

\[
\begin{split}
df
&= d\left(\frac{ \sqrt{3^{2}+x ^{2}} }{ 4 } +\frac{ 12-x }{ 6 }\right)\\
&= d\left(\frac{ \sqrt{3^{2}+x ^{2}} }{ 4 } \right)+d\left(\frac{ 12-x }{ 6 }\right)\\
&= \frac{d( \sqrt{3^{2}+x ^{2}}) }{ 4 } +\frac{ d(12-x) }{ 6 }\\

\end{split}
\]

Answer 10

\[
d(12-x) = -dx
\]

Answer 11

I'm assuming your equation is right, but lets see you're solving for time, so t = d/v, so your first distance looks right, your speed is 4miles/hr so that looks correct as well, (12-x)/6, nice you did it right :)

Answer 12

\[f'(x) \frac{ 1 }{ 8 } (3^2+x^2)^{-\frac{ 1 }{ 2 }} \times (9+2x) - 1\]

Answer 13

\[
\begin{split}
d(\sqrt{3^2+x^2})
&= d(\sqrt{u}) &\text{let } u = 3^2+x^2\\
&= \frac{1}{2\sqrt u}du \\
&= \frac{1}{2\sqrt{3^2+x^2}}d(3^2+x^2) \\
&= \frac{1}{2\sqrt{3^2+x^2}}d(x^2) \\
&= \frac{1}{2\sqrt{3^2+x^2}}(2x)dx \\

\end{split}
\]

Answer 14

Is my derivative correct? Do I just solve for 0 now?

Answer 15

\[
df = \left(\frac{x}{4\sqrt{3^2+x^2}}-\frac16\right)dx
\]

Answer 16

You made a few errors.

Answer 17

First of all... \[
d(3^2+x^2)\neq (9+2x)dx
\]

Answer 18

Yes, I see it now. Its 6+2x

Answer 19

No. \[
d(3^2) = 0
\]

Answer 20

\[
d(3^2+x^2) = d(3^2)+d(x^2) = 0 + 2x\;dx
\]

Answer 21

Ohhhh wow, thank you for clarifying that

Answer 22

\[f'(x) = \frac{ 1 }{ 8 }(9+x^2)^{-1/2} \times 2x-1/6 \implies \frac{ x }{ 4\sqrt{9+x^2} }-1/6\] so yeah you should get this without all the fancy notation ;).

Answer 23

Ok. I'm really learning a lot. Thank you. So now I just solve for 0.

Answer 24

Yes let f'(x) = 0, so \[\frac{ x }{ 4\sqrt{9+x^2} }-1/6 = 0\] should be easy peasy to solve from here :).

Answer 25

\[\frac{ 1 }{ 6 } = \frac{ x }{ 4\sqrt{(3^2+x^2)} }\]

\[4\sqrt{(3^2+x^2)} =\frac{ x }{ 6 }\]

\[\sqrt{3^2+x^2} =\frac{ 4x }{ 6 }\]

Is raising both sides to the exponent 2 to get rid of the the square root right?
\[3^2 + x^2 =\frac{ 16x }{ 36 } \]

Answer 26

Mhm I think you did something wrong here.

Answer 27

x/6?

Answer 28

Alright, lets do this step by step :)

Answer 29

Yes, you're are right!

Answer 30

I solved it :)

Answer 31

\[\frac{ x }{ 4\sqrt{9+x^2} } - \frac{ 1 }{ 6 } = 0\]

Answer 32

Oh ok :)

Answer 33

Thank you so much!!!!

Answer 34

Just to make sure what did you get :)

Answer 35

It was easy to make mistakes here, since it gets messy pretty quickly, but you seem to have gotten it, good job!

Answer 36

\[\frac{ 4\sqrt{9+x^2} }{ 3 } =x\]
\[\frac{ 4(9+x^2) }{ 9 } =x^2\]

\[\sqrt{\frac{ 36 }{ 5}} =x\]

Answer 37

x = 2.68

Answer 38

Hahah solving that was amazing :D

Answer 39

Yeah! and squareoot 36 = 6 so a little simplification there, but well done friend, where is your fields medal? :O

Answer 40

fields medal?

Answer 41

It's a medal the top mathematicians get in the world and it only comes around once every 4 years, and the way you do math, I say you deserve 5! :) hehe.

Answer 42

Hahaha XD you are too kind :))))))) You deserve 5^2 :D

Answer 43

Thank you again!

Answer 44

Haha, well good job friend, and take care!