Answer:
See Below.
Step-by-step explanation:
We have the equation:
[tex]\displaystyle y = \left(3e^{2x}-4x+1\right)^{{}^1\! / \! {}_2}[/tex]
And we want to show that:
[tex]\displaystyle y \frac{d^2y }{dx^2} + \left(\frac{dy}{dx}\right) ^2 = 6e^{2x}[/tex]
Instead of differentiating directly, we can first square both sides:
[tex]\displaystyle y^2 = 3e^{2x} -4x + 1[/tex]
We can find the first derivative through implicit differentiation:
[tex]\displaystyle 2y \frac{dy}{dx} = 6e^{2x} -4[/tex]
Hence:
[tex]\displaystyle \frac{dy}{dx} = \frac{3e^{2x} -2}{y}[/tex]
And we can find the second derivative by using the quotient rule:
[tex]\displaystyle \begin{aligned}\frac{d^2y}{dx^2} & = \frac{(3e^{2x}-2)'(y)-(3e^{2x}-2)(y)'}{(y)^2}\\ \\ &= \frac{6ye^{2x}-\left(3e^{2x}-2\right)\left(\dfrac{dy}{dx}\right)}{y^2} \\ \\ &=\frac{6ye^{2x} -\left(3e^{2x} -2\right)\left(\dfrac{3e^{2x}-2}{y}\right)}{y^2}\\ \\ &=\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\end{aligned}[/tex]
Substitute:
[tex]\displaystyle y\left(\frac{6y^2e^{2x}-\left(3e^{2x}-2\right)^2}{y^3}\right) + \left(\frac{3e^{2x}-2}{y}\right)^2 =6e^{2x}[/tex]
Simplify:
[tex]\displaystyle \frac{6y^2e^{2x}- \left(3e^{2x} -2\right)^2}{y^2} + \frac{\left(3e^{2x}-2\right)^2}{y^2}= 6e^{2x}[/tex]
Combine fractions:
[tex]\displaystyle \frac{\left(6y^2e^{2x}-\left(3e^{2x} - 2\right)^2\right) +\left(\left(3e^{2x}-2\right)^2\right)}{y^2} = 6e^{2x}[/tex]
Simplify:
[tex]\displaystyle \frac{6y^2e^{2x}}{y^2} = 6e^{2x}[/tex]
Simplify:
[tex]6e^{2x} \stackrel{\checkmark}{=} 6e^{2x}[/tex]
Q.E.D.
