Let `y=x^3-8x+7a n dx=f(t) and (dy)/(dx)=2a n dx=3 at t=0,` then find the value of `(dx)/(dt) at t = 0 `

Explanation

`"Let "y=x^(3)-8x+7`
`therefore" "(dy)/(dx)=3x^(2)-8`
It is given that when t=0, x=3. Therefore, when t=0,
`(dy)/(dx)=3xx3^(2)-8=19`
`"Also ", (dy)/(dx)=(dy//dt)/(dx//dt)`
Since when `t=0, (dy)/(dx)=19 and (dy)/(dt) = 2, from (1)`
`19=(2)/(dx//dt) or (dx)/(dt)=(2)/(19)`