What is the derivative of (x^2-2xe^{4x}) ?

The derivative of (x^2-2xe^{4x}) is 4e^{4x}x^2-6e^{4x}x-2e^{4x}

What is the first derivative of (x^2-2xe^{4x}) ?

The first derivative of (x^2-2xe^{4x}) is 4e^{4x}x^2-6e^{4x}x-2e^{4x}

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

4 e^{4 x} x^{2} - 6 e^{4 x} x - 2 e^{4 x}

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2} - 2 x, g = e^{4 x}

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

\frac{d}{d x} (x^{2} - 2 x) = 2 x - 2

\frac{d}{d x} (e^{4 x}) = e^{4 x} \cdot 4

= (2 x - 2) e^{4 x} + e^{4 x} \cdot 4 (x^{2} - 2 x)

Simplify

(2 x - 2) e^{4 x} + e^{4 x} \cdot 4 (x^{2} - 2 x) : 4 e^{4 x} x^{2} - 6 e^{4 x} x - 2 e^{4 x}

= 4 e^{4 x} x^{2} - 6 e^{4 x} x - 2 e^{4 x}

\int (x + 1)^{- 8} d x

\frac{\partial}{\partial y} (x^{2} - 2 x y + y^{2})

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

\int x x^{2} + a^{2} d x

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

What is the derivative of (x^2-2xe^{4x}) ?
The derivative of (x^2-2xe^{4x}) is 4e^{4x}x^2-6e^{4x}x-2e^{4x}
What is the first derivative of (x^2-2xe^{4x}) ?
The first derivative of (x^2-2xe^{4x}) is 4e^{4x}x^2-6e^{4x}x-2e^{4x}