Find dy/dx, given y= sqrt(3x^2+4)?

Expert Answers
hala718 eNotes educator| Certified Educator

y = sqrt (3x^2 +4)

sinnce the sqrt has more that one factor then we need to use the chain rule

assume 3x^2+ 4 = g(x)= t

==> y = sqrt g(x)= sqrt t

==> dy/dx = dy/dt * dt/dx

howeverr dy/dt = (sqrt t) '= 1/(2sqrt t) = 1/[(2sqrt(3x^2+4)]

   and, dt/dx = (3x^2+ 4)' = 6x

then dy/dx = 1/[2 qrt (3x^2+4)] * 6x

               = 3x / sqrt (3x^2+4)

giorgiana1976 | Student

We'll use the substitution method and the chain rule, so that if we'll note 3x^2+4 = t, we'll get:

dy/dx = (dy/dt)(dt/dx)

dy/dt = (sqrt t)'=1/ 2sqrt t

dt/dx = (3x^2+4)' = 3*2*x^(2-1) = 6x

dy/dx = (dy/dt)(dt/dx) = (1/ 2sqrt t)*(6x)

But t = 3x^2+4, so dy/dx = (1/ 2sqrt(3x^2+4))*(6x) 

dy/dx = 6x/ 2sqrt(3x^2+4)

After simplyfying:

dy/dx = 3x/sqrt(3x^2+4)

neela | Student

y = sqrt(3x^2+4)

To find dy/dx.

Solution:

y = (3x^2+4)^(1/2).

We use the cchain rule method  where y = u(v(x)).

Then dy/dx = {u(v(x))}' = {u(v)]' *[v(x)]'

So  dy/dx =  {(3x^2+4)^(1/2-1)}{3x^2+4}'

=(1/2)(3x^2+4)[3*2x^(1/2-1)+0]

= (1/2)(3x^2+4)^(-1/2) ](6x)

= 6x/sqrt(3x^2+4)