Question

Use the chain rule, in Leibniz notation, to find dy/dx
y= u(u^2 +3)^3 , u= (x+3)^2 , x = -2

Answer 1

\[\frac{dy}{du}=(u^2+3)^3+3u(u^2+3)^2\times 2u\]
\[\frac{dy}{du}=(u^2+3)^3+6u^2(u^2+3)^2\] by the chain rule and the product rule

Answer 2

yes i got this too

Answer 3

you can also write this as
\[(u^2+3)^2 (7 u^2+3)\] if you like

Answer 4

\[u=(x+3)^2\]
\[\frac{du}{dx}=2(x+3)=2x+6\]

Answer 5

multiply together, and replace x by -2

Answer 6

umm,that's exactly what i did...but somehow i got a really big number..

Answer 7

looks like it would be really big to me

Answer 8

cuz u = (x+3)^2 , when i sub it in, it will be （(x+3)^4 +3 )^2

well, the answer is only 540 or so

Answer 9

it's 320

Answer 10

ok if x = -2, then
\[ u = (-2+3)^2=1\] so maybe not so big

Answer 11

\[(u^2+3)^2 (7 u^2+3)\]
\[4^2(10)=160\] then multiply by
\[2(-2)96=-2\]

Answer 12

sorry i mean
\[2\times -2+6=2\]

Answer 13

oh..so u sub in x first before multiplying??
then i think i've made a mistake when i combine both equations! ok thxx i will do it ur way then :)

Answer 14

i get 320 if i am not mistaken

Answer 15

yeah that is way easier