What is the solution for derivative of x^2e^2+2^y=5 ?

The solution for derivative of x^2e^2+2^y=5 is y=(ln(5x+c_{1}-x^2e^2))/(ln(2))

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derivative of x^2e^2+2^y=5

\frac{d}{d x} (x^{2} e^{2} + 2^{y}) = 5

y = \frac{ln ( 5 x + c _{1} - x ^{2} e ^{2} )}{ln ( 2 )}

Solution steps

\frac{d}{d x} (x^{2} e^{2} + 2^{y}) = 5

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

x^{2} e^{2} + 2^{y} = \int 5 d x

\int 5 d x = 5 x + c_{1}

x^{2} e^{2} + 2^{y} = 5 x + c_{1}

Isolate

y : y = \frac{ln ( 5 x + c _{1} - x ^{2} e ^{2} )}{ln ( 2 )}

y = \frac{ln ( 5 x + c _{1} - x ^{2} e ^{2} )}{ln ( 2 )}

4 x^{4} - 4 x^{2}, x^{2}

\frac{d}{d x} (\frac{x - 1}{2})

\frac{d}{d x} (\frac{2 x - 3}{( x ^{2} + 4 ) ^{2}})

\frac{d}{d x} (- x^{2} + 4 x + 1)

\int \frac{( 4 x ^{2} + 4 )}{( x ^{2} - 2 x + 2 ) ^{2}} d x

What is the solution for derivative of x^2e^2+2^y=5 ?
The solution for derivative of x^2e^2+2^y=5 is y=(ln(5x+c_{1}-x^2e^2))/(ln(2))