Question

Calc (derivatives) the question is y=x-6-(5/x)-(x/6)+e^(5x)
I got 3+e^(5x), but I don't think that is right. I took the derivative of (-5/x)^(1), which would be 1, but I am kind of lost as to what to do. Could I instead do -5x^(-1)?

Answer 1

no it is not right

Answer 2

does it really say
\[\large y=x-6-\frac{5}{x}-\frac{x}{6}+e^{5x}\]?

Answer 3

like this ?
\(\Huge y=x-6-\frac{5}{x}-\frac{x}{6}+e^{5x}\)

Answer 4

yes

Answer 5

yes

Answer 6

\[\frac{ d }{ dx }\left( \frac{ u }{v } \right)=\frac{ vu'-uv' }{v^2 }\]

Answer 7

What I am doing now, after restarting, is (just for the fractions) -5(x^(-1), which means i have to use the multplication rule

Answer 8

ok first of all
\[x-\frac{x}{6}=\frac{5}{6}x\]

Answer 9

and the derivative of \(\frac{5}{6}x\) is \(\frac{5}{6}\)

Answer 10

or\[\frac{ d }{ dx }\left( \frac{ 1 }{x^n } \right)=\frac{ d }{dx }\left( x ^{-n} \right)=-nx ^{-n-1}\]

Answer 11

the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\) so the derivative of \(-\frac{5}{x}\) is \(\frac{5}{x^2}\)

Answer 12

use no rule
use your memory
the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\) and by the chain rule, the derivative of \(\frac{1}{f(x)}\) is \(-\frac{f'(x)}{f^2(x)}\)

Answer 13

My teacher wants us to show our work. How did you get -5/x =5/(x^(2)?

Answer 14

explanation above
if showing work is really necessary, either you have to show
\[\lim_{h\to 0}\frac{\frac{-5}{x+h}+\frac{5}{x}}{h}=\frac{5}{x^2}\] which is the only work to do
if you get to use shortcut rules, then the quickest shortcut is to remember it
the reciprocal is a very common function, and the derivative never changes

Answer 15

wait, I think I have it. is the answer:\[-5+\frac{ 5 }{ x^2 }+\frac{ x }{ 6^2 } +e ^{5x}\]

Answer 16

there are two you should simply remember without using any rules
one is the rule for square roots
\[\frac{d}{dx}\sqrt{x}]=\frac{1}{2\sqrt{x}}\] and by the chain rule
\[\frac{d}{dx}[\sqrt{f(x)}]=\frac{f'(x)}{2\sqrt{f(x)}}\]
those should be immediate

Answer 17

no combine like terms first

Answer 18

\[-\frac{x}{6}=-\frac{1}{6}x\]

Answer 19

and so the derivative of \[-\frac{1}{6}x\] is \[-\frac{1}{6}\]

Answer 20

also, by the chain rule, the derivative of \(e^{5x}\) is
\[e^{5x}\frac{d}{dx}[5x]\] i.e.
\[5e^{5x}\]

Answer 21

oh, I see

Answer 22

I thought that the derivative of e^(5x) was just e^(5x)

Answer 23

final answer should be
\[\frac{5}{6}+\frac{5}{x^2}+5e^{5x}\] i think
the
\[x-\frac{x}{6}\] was a trick to make sure you were paying attention i guess
most times you do not see a function without the like terms combined

Answer 24

chain rule for these, and it is easy to forget when you have things like
\[\sin(2x)\] for example
the derivative of
\[\sin(2x)\] is
\[2\cos(2x)\] don't forget the 2 out front

Answer 25

Ok, I think i am getting it.

Answer 26

I got it, thanks!!!