if you differesiate 1=2ax dx/dt+b dx/dt. then you common dx/dt.so your equation will be made1/(2ax+b) =v.then again differensiate.A = 2av(-1)/(2ax+b)^2

                                      v^2 =1/(2ax+b)^2.put upper equation .you can get answer -2a v^3

ax²+bx=t

Acceleration is double differntiation or derivative for the relation displacement and time given:

thus, dx/dt =>  (2ax + b )dx = dt

now dv/dt = 2a thus, acc. is ''2a''

plz approve!

 Just do dx/dt , you will get v then repeat the step by doing dv/dt , you will get acceleration.

dx/dt = 2ax + b

dv/dt = 2a + 0 = 2a

that''s the answer.

A = dv/dt = d²x/dt²

where A is the acceleration.

By finding the second derivative of the given expression, we will find the acceleration.

t = ax² + bx

1 = 2ax dx/dt + b dx/dt

dx/dt = 1/(2ax+b)

d²x/dt² = -2a/(2ax+b)²

A = d²x/dt² = -2a/(2ax+b)²

Differentiating with respect to time:

1=2ax(dx/dt) + b(dx/dt)

=> v=1/(2ax + b)

=>a=0.(Differentiating wrt time, no t in RHS)

Therfore, a=0.

2a

dx/dt = 2ax + b

dv/dt = 2a + 0 = 2a

plz approve..!!

the given expression is a quadratic in x.

to get the acceleration we need to find x as a function of t.

solving the quadratic gives x = [{b² + 4at}^1/2 - b]/2a . 

differentiating twice with respect to t we get the acceleration as -2a*{b² + 4at}^-3/2.

dt/dx = 2ax +b dx/dt = 1/2ax+bV=1/2ax+b. .....(I)again differentiate,A=2av(-1)/(2ax+b)^2Put value of v from. eq 1 Final answer : -2av^3