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

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

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

Answer

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}})