Question

Hi. So, I've been having difficulty with the chain rule for the past week. I haven't asked my teacher for help because his way of teaching is extremely confusing and not helpful at all. I would appreciate if someone could walk me through the problem below. Thank you.

Answer 1

post your problem

Answer 2

here is the problem

Answer 3

i can't read properly

Answer 4

is it \[\frac{ 1 }{ 2 }x^2\sqrt{16-x^2}\] ?

Answer 5

is it correct?

Answer 6

are you here?

Answer 7

it equals y. but other than that, it is correct

Answer 8

you want to find dy/dx.

Answer 9

i will be back after dinner.wait for few minutes.

Answer 10

okay, I will be here

Answer 11

\[\frac{ dy }{ uv }=uv \prime+u \prime v\]
where u and v are functions of x

Answer 12

\[\frac{ dy }{ dx }=\frac{ d }{ dx }\left[ \frac{ 1 }{ 2 }x^2\left( 16-x^2 \right)^{\frac{ 1 }{ 2 }} \right]\]
\[=\frac{ 1 }{ 2 }\left[ x^2 \times \frac{ 1 }{ 2 }\left( 16-x^2 \right)^{\frac{ 1 }{ 2 }-1}\left(0 -2x \right)+2x \left( 16-x^2 \right)^{\frac{ 1 }{ 2 }} \right]\]
\[=\frac{ 1 }{ 2 }\left[ \frac{ -x^3 }{ \sqrt{16-x^2} }+2x \sqrt{16-x^2} \right]\]
\[=\frac{ 1 }{ 2 }\left[ \frac{-x^3+2x \left( 16-x^2 \right) }{ \sqrt{16-x^2} } \right]\]
\[=\frac{ 1 }{ 2 }\left[ \frac{ -x^3+32x-2x^3 }{ \sqrt{16-x^2} } \right]\]
\[=\frac{ 32x-3 x^3 }{2 \sqrt{16-x^2} }\]

Answer 13

i am leaving now

Answer 14

Thank you for your help

Answer 15

you have any questions.