What is the solution for derivative of f(x)=8^{(x^2+4x)} ?

The solution for derivative of f(x)=8^{(x^2+4x)} is f(x)=integral of 8^{x^2+4x}

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derivative of f(x)=8^{(x^2+4x)}

\frac{d}{d x} f (x) = 8^{(x^{2} + 4 x)}

f (x) = \int 8^{x^{2} + 4 x} d x

Solution steps

\frac{df}{d x} = 8^{(x^{2} + 4 x)}

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

then

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

f (x) = \int 8^{x^{2} + 4 x} d x

f (x) = \int 8^{x^{2} + 4 x} d x

\frac{d y}{d x} = e^{2 x + 3 y}, y (0) = 0

\int_{6}^{\infty} x e^{- 2 x} d x

\int - 5 cos (x) d x

y^{^{''}} - 8 y^{^{'}} + 10 y = 0

\int 8 sec^{2} (x) d x

What is the solution for derivative of f(x)=8^{(x^2+4x)} ?
The solution for derivative of f(x)=8^{(x^2+4x)} is f(x)=integral of 8^{x^2+4x}