Question

Help with 30 please. thanks.​

Answer 1

See Below.

Explanation:

We have the equation:

`\displaystyle y = \left(3e^(2x)-4x+1\right)^{{}^1\! / \! {}_2}`

And we want to show that:

`\displaystyle y (d^2y )/(dx^2) + \left((dy)/(dx)\right) ^2 = 6e^(2x)`

Instead of differentiating directly, we can first square both sides:

`\displaystyle y^2 = 3e^(2x) -4x + 1`

We can find the first derivative through implicit differentiation:

`\displaystyle 2y (dy)/(dx) = 6e^(2x) -4`

Hence:

`\displaystyle (dy)/(dx) = (3e^(2x) -2)/(y)`

And we can find the second derivative by using the quotient rule:

`\displaystyle \begin{aligned}(d^2y)/(dx^2) & = ((3e^(2x)-2)'(y)-(3e^(2x)-2)(y)')/((y)^2)\\ \\ &= (6ye^(2x)-\left(3e^(2x)-2\right)\left((dy)/(dx)\right))/(y^2) \\ \\ &=(6ye^(2x) -\left(3e^(2x) -2\right)\left((3e^(2x)-2)/(y)\right))/(y^2)\\ \\ &=(6y^2e^(2x)-\left(3e^(2x)-2\right)^2)/(y^3)\end{aligned}`

Substitute:

`\displaystyle y\left((6y^2e^(2x)-\left(3e^(2x)-2\right)^2)/(y^3)\right) + \left((3e^(2x)-2)/(y)\right)^2 =6e^(2x)`

Simplify:

`\displaystyle (6y^2e^(2x)- \left(3e^(2x) -2\right)^2)/(y^2) + (\left(3e^(2x)-2\right)^2)/(y^2)= 6e^(2x)`

Combine fractions:

`\displaystyle (\left(6y^2e^(2x)-\left(3e^(2x) - 2\right)^2\right) +\left(\left(3e^(2x)-2\right)^2\right))/(y^2) = 6e^(2x)`

Simplify:

`\displaystyle (6y^2e^(2x))/(y^2) = 6e^(2x)`

Simplify:

`6e^(2x) \stackrel{\checkmark}{=} 6e^(2x)`

Q.E.D.