Question

Need clarification please
Topic: Related rates of change- Differentiation

show that:
\[\frac{ d }{ dx }(\frac{ v^2 }{ 2 })=\frac{ d^2x }{ dt^2 }\]
Where x represents the distance travelled in t seconds by a particle moving with velocity v

This is what i have so far:

Answer 1

\[\frac{ dx }{ dt }=v\]
Differentiate this with respect to time and get

\[\frac{ d^2x }{ dt^2 }=\frac{ dv }{ dt } \]

Answer 2

we know:
\[\frac{ dv }{ dt }=\frac{ dv }{ dx }\times \frac{ dx }{ dt }\]

Answer 3

which is
\[\frac{ dv }{ dt }=\frac{ dv }{ dx } \times v\]

Answer 4

im lost now

Answer 5

By the chain rule, the left hand side is
\[\frac{d}{dx}\left[\frac{v^2}{2}\right]=\frac{1}{2}\left(2v\frac{dv}{dx}\right)=v\frac{dv}{dx}\]which you showed is equal to \(\dfrac{dv}{dt}\).

Answer 6

I am not sure how you for the left side of the equation you wrote above

Answer 7

Do you know what the chain rule is?

Answer 8

i think the v and t are confusing me because i keep writing x and y. i might try doing it again