Question

find \\( \frac { d ^ { 2 } y } { d x ^ { 2 } } \\).
\\( e ^ { 2 y } + x = 8 y \\)
\\( \frac { d ^ { 2 } y } { d x ^ { 2 } } = \\)

Explanation:

Step1: Differentiate \(e^{2y}+x = 8y\) with respect to \(x\)

Using the chain rule \((e^{u})^\prime=e^{u}\cdot u^\prime\) (where \(u = 2y\)), we have:
\(e^{2y}\cdot2y^\prime+1=8y^\prime\)
Rearrange to solve for \(y^\prime\):
\(2e^{2y}y^\prime - 8y^\prime=-1\)
\(y^\prime(2e^{2y}-8)=-1\)
\(y^\prime=\frac{-1}{2e^{2y}-8}=\frac{1}{8 - 2e^{2y}}\)

Step2: Differentiate \(y^\prime=\frac{1}{8 - 2e^{2y}}\) with respect to \(x\)

Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = 1\), \(u^\prime=0\), \(v = 8 - 2e^{2y}\), and \(v^\prime=-4e^{2y}y^\prime\)
\(y^{\prime\prime}=\frac{0\cdot(8 - 2e^{2y})-1\cdot(- 4e^{2y}y^\prime)}{(8 - 2e^{2y})^{2}}\)
Substitute \(y^\prime=\frac{1}{8 - 2e^{2y}}\) into the above formula:
\(y^{\prime\prime}=\frac{4e^{2y}\cdot\frac{1}{8 - 2e^{2y}}}{(8 - 2e^{2y})^{2}}\)
Simplify the expression:
\(y^{\prime\prime}=\frac{4e^{2y}}{(8 - 2e^{2y})^{3}}\)

Answer:

\(\frac{4e^{2y}}{(8 - 2e^{2y})^{3}}\)