What is the derivative of xe^{4x+1}+1/2 ?

The derivative of xe^{4x+1}+1/2 is 4e^{4x+1}x+e^{4x+1}

What is the first derivative of xe^{4x+1}+1/2 ?

The first derivative of xe^{4x+1}+1/2 is 4e^{4x+1}x+e^{4x+1}

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

4 e^{4 x + 1} x + e^{4 x + 1}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (x e^{4 x + 1}) = e^{4 x + 1} + 4 e^{4 x + 1} x

\frac{d}{d x} (\frac{1}{2}) = 0

= e^{4 x + 1} + 4 e^{4 x + 1} x + 0

Simplify

= 4 e^{4 x + 1} x + e^{4 x + 1}

f (x) = x^{2} - sec (x)

x y^{^{''}} = y^{^{'}}

s (t) = \frac{400 t}{( t ^{2 + 5} ) ^{3}}

x \to 4 lim (15)

\frac{d y}{d x} = \frac{( x + y )}{6}

What is the derivative of xe^{4x+1}+1/2 ?
The derivative of xe^{4x+1}+1/2 is 4e^{4x+1}x+e^{4x+1}
What is the first derivative of xe^{4x+1}+1/2 ?
The first derivative of xe^{4x+1}+1/2 is 4e^{4x+1}x+e^{4x+1}