Question

find the critical numbers of y=x/(x^2+16)

Answer 1

Find the derivative and equate it to 0.

Answer 2

Not sure how to find the derivative

Answer 3

Could you please explain how I would solve for the Derivative?

Answer 4

are you in algebra or calculus class? @sharrisf

Answer 5

calculus @mark_o.

Answer 6

Differntial Calculus.... u have to use quotient Rule....

Answer 7

ah ok so you know how to find slope of the tangent? you can do it here
\[\lim (x \rightarrow0) \frac{ \Delta y }{ \Delta x }=\lim ( x \rightarrow0) \frac{ f(x+h)-f(x) }{ h } \]

Answer 8

???? let me help you,what do you think? i need your out put..lol

Answer 9

Let's put it this way, ever since we started with limits I've been hearing greek and seeing squiggles

Answer 10

i understand that we all started there..lol

Answer 11

\[y+\Delta y=f(x+\Delta x)\] then
\[\Delta y=f(x+\Delta x) -y\]

Answer 12

if y=16x^2 then
\[y+\Delta y=16(x+\Delta x)^{2}\]

Answer 13

therefore
\[\Delta y=16(x+\Delta x)^{2}-y\] , but y=16x^2 so subs it to the eq

Answer 14

its now
\[\Delta y=16(x+\Delta x)^{2}-16x ^{2}\]

Answer 15

you can now expand that to
\[\Delta y=16(x ^{2}+2x \Delta x +\Delta x ^{2})-16x ^{2}\]
expand that more

Answer 16

\[\Delta y=16x ^{2}+32x \Delta x+16\Delta x ^{2}-16x ^{2}\]
do addition or subtraction what ever needed

Answer 17

so
\[\Delta y=32x \Delta x+16\Delta x ^{2}\]
thats your delta y or your changed in y

Answer 18

also
\[\frac{ \Delta y }{ \Delta x }=\frac{ 32x \Delta x +16\Delta x ^{2} }{ \Delta x }\]
you can divide them

Answer 19

therefore
\[\frac{ \Delta y }{ \Delta x }=\frac{ 32x +16\Delta x }{ \Delta x }\]

Answer 20

now apply the limit here as x approach to zero
\[ \lim_{\Delta x \rightarrow 0}\frac{ \Delta y }{ \Delta x }=32x\]