Question

find dy/dx of (x+y)^3=x^3+y^3

Answer 1

the process is pretty simple with the derivative rules .. what is the issue you are having?

Answer 2

the answer im getting doesnt match the back of the book
i got dy/dx=[3x ^{2}-3x-3y \div3x+3y-3y ^{2}\]

Answer 3

can you step me thru it so I can see where you may be going astray at?

Answer 4

it may just be a matter of simplification as well

Answer 5

ok heres wht i did

Answer 6

also if u could help me with this one, i thought i did it but once again my answer doesnt match the back of the books. \[y ^{2}=x ^{2}-49divx ^{2}+49\]

Answer 7

\[(x+y)^3 = x^3+y^3\]
\[3(x+y)^2~(x+y)' = 3x^2x'+3y^2y'\]
\[3(x+y)^2~(x'+y') = 3x^2x'+3y^2y'\]
\[x'=\frac{dx}{dx}=1\]
\[3(x+y)^2~(1+y') = 3x^2+3y^2y'\]
\[(x+y)^2+y'(x+y)^2 = x^2+y^2y'\]
\[y'(x+y)^2-y^2y' = x^2-(x+y)^2\]
\[y'[(x+y)^2-y^2] = x^2-(x+y)^2\]
\[y' = \frac{x^2-(x+y)^2}{(x+y)^2-y^2}\]
\[y' = \frac{[x-(x+y)]~[x+(x+y)]}{[(x+y)+y]~[(x+y)-y]}\]
\[y' = \frac{[-y]~[2x+y]}{[x+2y]~[x]}\]
doesnt look to be any common factors to pull out

Answer 8

i think you ignored a ^2 at the start of it

Answer 9

duh. yes i forgot that squared. thnk you

Answer 10

wait im confused as to how all the numbers got cancled out?

Answer 11

\[y ^{2}=\frac{x ^{2}-49}{x ^{2}+49}\]
\[y ^{2}=(x ^{2}-49)(x ^{2}+49)^{-1}\]
now its just the product rule
\[2y~y'=2xx'(x ^{2}+49)^{-1}-2xx'(x ^{2}-49)(x ^{2}+49)^{-2}\]
divide out the 2 and let x' = 1
\[y~y'=x(x ^{2}+49)^{-1}-x(x ^{2}-49)(x ^{2}+49)^{-2}\]
we can factor out that (x^2+49)^2 if need be
\[y~y'=\frac{x(x ^{2}+49)-x(x ^{2}-49)}{(x ^{2}+49)^{2}}\]
\[y'=\frac{x(x ^{2}+49)-x(x ^{2}-49)}{y(x ^{2}+49)^{2}}\]
\[y'=\frac{x[(x ^{2}+49)-(x ^{2}-49)]}{y(x ^{2}+49)^{2}}\]
\[y'=\frac{x(49+49)}{y(x ^{2}+49)^{2}}\]

Answer 12

3 is a common factor to all the terms; dividing both sides by 3 cancels them out

Answer 13

\[\frac{\cancel3(x+y)^2~(x'+y') = \cancel3x^2x'+\cancel3y^2y'}{\cancel3}\]
\[(x+y)^2~(x'+y') = x^2x'+y^2y'\]

Answer 14

what does the book give us?

Answer 15

ok i got it now.
and the book gives the same answer you got

Answer 16

yay!!

Answer 17

now im working on the second problem i gave you
could u work it through using the quotient rule, cuz thats how i did it but didnt get the right answer

Answer 18

i despise the quotient rule ... its just got to many moving parts to me.

a qoutient is just another form of a product, and the product rule is sooo much easier to keep track of to me

Answer 19

actually i think i found where i messed up, i forgot the bottom half of the quotient rule. let me work it thru one mroe time

Answer 20

\[\frac tb=tb^{-1}\]
\[[\frac tb]'=\frac{bt'-b't}{b^2}\]
\[[tb^{-1}]'=t'b^{-1}+tb'^{-1}\]
\[[tb^{-1}]'=t'b^{-1}-tb'b^{-2}\]
\[[tb^{-1}]'=\frac{1}{b^2}(t'b-tb')\]
\[[tb^{-1}]'=\frac{t'b-tb'}{b^2}\]

Answer 21

ok im stuck on the making of the produuct rule. i got everything up until the taking the derviative of (x^2+49)^-1

Answer 22

would we use the chain rule as well?

Answer 23

power rule, drop the -1, and subtract 1
chain rule pops out the derivative of the innards