Question

f(x)=x(8-x^2)^1/2
find where the first derivative is zero or undefined

Answer 1

you got the derivative?

Answer 2

no I'm stuck on it

Answer 3

i take it it is
\[f(x)=x\sqrt{8-x^2}\] use the product rule and chain rule to get
\[f'(x)=\sqrt{8-x^2}-\frac{x^2}{\sqrt{8-x^2}}\]

Answer 4

you need it step by step?

Answer 5

yes, please

Answer 6

k
it is a product right \(x\times \sqrt{8-x^2}\) therefore you need the product rule
\[(fg)'=f'g+g'f\]

Answer 7

here we could say \(f(x)=x, f'(x)=1\) and \(g(x)=\sqrt{8-x^2}\)

Answer 8

now we need \(g'(x)\) which requires the chain rule, or maybe just knowing that the derivative of the square root of something is the derivative of something over 2 root of something
in this case you get \[g'(x)=\frac{-2x}{2\sqrt{8-x^2}}\]

Answer 9

Im confused on the last message

Answer 10

ok lets do a quick example
what is the derivative of \(\sqrt{\sin(x)}\)? it is
\[\frac{\cos(x)}{2\sqrt{\sin(x)}}\]

Answer 11

ok

Answer 12

what is the derivative of \(\sqrt{x^2+3x}\) ? it is
\[\frac{2x+3}{2\sqrt{x^2+3x}}\]

Answer 13

and similarly the derivative of \(\sqrt{8-x^2}\) is \(\frac{-2x}{2\sqrt{8-x^2}}\) although you probably want to cancel the twos

Answer 14

plug all that stuff in to here \((fg)'=f'g+g'f\) and you get
\[\sqrt{8-x^2}-\frac{x^2}{\sqrt{8-x^2}}\]

Answer 15

in order to find the zeros you have to actually do the subtraction
to find where it is undefined is just a matter of setting \(8-x^2=0\) and solving for \(x\) pretty much in your head

Answer 16

don't we add in the product rule

Answer 17

that was the product rule

Answer 18

\[(fg)'=f'g+g'f\] with
\[f(x)=x,f'(x)=1,g(x)=\sqrt{8-x^2}, g'(x)=\frac{-x}{\sqrt{8-x^2}}\]

Answer 19

plug them directly in to the formula, and see that you get (i hope) the answer i wrote above

Answer 20

\[1 * \sqrt{8-x ^{2}} + 2x/2\sqrt{2-x^2} * x\] This?

Answer 21

yes except you missed the minus sign

Answer 22

the derivative of \(8-x^2\) is \(-2x\)

Answer 23

oh ok thats where that (-) came from

Answer 24

yup

Answer 25

now you have to actually do the subtraction to find the zeros

Answer 26

\[\sqrt{8-x ^{2}} - x^2 / \sqrt{8-x^2}\]
This is what I have now

Answer 27

should I find the common denominator?

Answer 28

? you only have one denominator

Answer 29

duh nevermind

Answer 30

what if I do then I can simplify and get rid of the first \[\sqrt{8-x^2}\]

Answer 31

it will go when you subtract

Answer 32

is the top where it is equal to zero and undefined on the bottom

Answer 33

\[\sqrt{8-x^2}-\frac{x^2}{\sqrt{8-x^2}}\]
\[=\frac{8-x^2-x^2}{\sqrt{8-x^2}}\]

Answer 34

better known as

\[\frac{8-2x^2}{\sqrt{8-x^2}}\]

Answer 35

clear how i got that?

Answer 36

yes
okay the top says that the first derivative is equal to 0 when x = + or - 2
but how do I mess with the radical

Answer 37

in the denominator? you don't
the denominator is zero when \(8-x^2=0\)

Answer 38

is that where it is undefined?

Answer 39

yes, when the denominator is zero the derivative is undefined
you get \(x=\pm\sqrt8\) pretty much in your head

Answer 40

ok thank you

Answer 41

not really that surprising, since those numbers are the endpoints of the domain of \(\sqrt{8-x^2}\)

Answer 42

yw
lotta algebra on this one huh?

Answer 43

yes and I stink at algebra lol and I always thought my weakness was geometry and trig
i have one more if you can help. I think its alot easier :)

Answer 44

sure

Answer 45

okay thanks I appreciate it quiz tomorrow on this stuff
I just have to find the first derivative of \[y=3^{2x} + 3e ^{2x}-\ln(3x^2)\]

Answer 46

ok lets go step by step, starting with the easy part

Answer 47

sounds good to me

Answer 48

the derivative of \(e^x\) is \(e^x\)
therefore by the chain rule the derivative of \(e^{2x}\) is \(2e^{2x}\) making the derivative of \(e^{2x}\) as \(6e^{2x}\)
so far so good?

Answer 49

oops typo
last line should be "making the derivative of \(3e^{2x}\) as \(6e^{2x}\)"

Answer 50

so far so good?

Answer 51

ok I got 6e^2x

Answer 52

ok now how about \(3^{2x}\)
or more generally \(b^x\) do you know the derivative of that?

Answer 53

just b

Answer 54

ooooh no!

Answer 55

lol

Answer 56

the derivative of \(b^x\) is \(b^x\ln(b)\)

Answer 57

for example the derivative of \(10^x\) is \(10^x\ln(10)\)

Answer 58

and so by the chain rule the derivative of \(3^{2x}\) is \(2\times \ln(3)\times 3^{2x}\)

Answer 59

ok

Answer 60

and finally we get to the easiest one \(-\ln(3x^2)\)
do you know the derivative of \(\ln(x)\) ?

Answer 61

1/x

Answer 62

ok so now you have a choice, you can either use the chain rule or rewrite the log, same answer in either case

Answer 63

\[\ln(3x^2)=\ln(3)+2\ln(x)\] and since \(\ln(3)\) is just some number, the derivative of \(-\ln(3x^2)\) is \(-\frac{2}{x}\) that is one way

Answer 64

or by the chain rule you get
\[-\frac{6x}{3x^2}\] which is also \(-\frac{2}{x}\)

Answer 65

ok so to clean it up a bit the first part 2 * ln3 * 3^2x how can I simplify that

Answer 66

you can't really
2 and ln(3) are just numbers
you can write it without the times signs, but that is all

Answer 67

ok thanks you have been super helpful I really appreciate it

Answer 68

yw

Answer 69

have a great night

Answer 70

good luck on your quiz!