Question

Y=t²+t³ and x=t-t⁴ the. Find d²y/dx²

Answer 1

Given :  y=t²+t³ and x=t-t⁴

To find : d²y/dx²

Solution:

y = t² + t³

=> dy/dt  = 2t + 3t²

x  = t  - t⁴

=> dx/dt  = 1  - 4t³

(dy/dt)/(dx/dt) =  (2t + 3t²)/(1  - 4t³)

=> dy/dx =   (2t + 3t²)/(1  - 4t³)

=>  d²y/dx² = ((2t + 3t²) (-1)/ (1  - 4t³)² ) (-12t² dt/dx)   + (1/(1  - 4t³))(2 + 6t)dt/dx

=>  d²y/dx² = ((dt/dx)/ (1  - 4t³)² )  ( 24t³ + 36t⁴ - 24t⁴ - 8t³ + 6t + 2 )

dx/dt  = 1  - 4t³  => dt/dx  = 1/(1  - 4t³)

=>   d²y/dx² =  ( 12t⁴ + 16t³ + 6t + 2) / (1  - 4t³)³

 d²y/dx² =  ( 12t⁴ + 16t³ + 6t + 2) / (1  - 4t³)³

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