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If y is a function of z and z = ax, then prove that d^2y/dx^2=a^2(d^2y/dx^2)
Let y=f(x) , dy/dx = f'(x) and d²y/dx²= f''(x)Z=axWe know dy/dz can also be written as(Dy/dx)÷(dz/dx)Now if we differentiate the numerator and denominatorWe get dy/dz=(dy/dx)÷a = f'(x)/aDifferentiating the the above equation againWe get ,d²y/dz² = [ d(f'(x))/dz * a - f'(x)*d(a)/dz ] ÷ a² (APPLYING QUOTIENT RULE)a²*d²y/dz² = d(f'(x))/dx *a *dx/dz - f'(x) *0 (APPLYING CHAIN RULE)a²*d²x/dz² = f''(x) *a* (dx / d(ax)). (Z=ax)a²*d²x/dz² = f''(x) *a*(dx/a * dx) (a is taken outside Being a constant , contsant rule)Cutting a and dx from numerator and denominator (i assume that a is contsant and is not equal to zero. This information is a must. Otherwise the whole question is itself invalid)We get our final result asa²*d²x/dz² = f''(x)= a² * d²x/dz² = d²y/dx²Hence we have proved the given question!
Let y=f(x) , dy/dx = f'(x) and d²y/dx²= f''(x)
Z=ax
We know dy/dz can also be written as
(Dy/dx)÷(dz/dx)
Now if we differentiate the numerator and denominator
We get dy/dz=(dy/dx)÷a = f'(x)/a
Differentiating the the above equation again
We get ,
d²y/dz² = [ d(f'(x))/dz * a - f'(x)*d(a)/dz ] ÷ a² (APPLYING QUOTIENT RULE)
a²*d²y/dz² = d(f'(x))/dx *a *dx/dz - f'(x) *0 (APPLYING CHAIN RULE)
a²*d²x/dz² = f''(x) *a* (dx / d(ax)). (Z=ax)
a²*d²x/dz² = f''(x) *a*(dx/a * dx) (a is taken outside Being a constant , contsant rule)
Cutting a and dx from numerator and denominator (i assume that a is contsant and is not equal to zero. This information is a must. Otherwise the whole question is itself invalid)
We get our final result as
a²*d²x/dz² = f''(x)
= a² * d²x/dz² = d²y/dx²
Hence we have proved the given question!
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