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

The derivative of log_{4}(xe^x) is 1/(2ln(2)x)+log_{4}(e)

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

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

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\frac{d}{d x} (lo g_{4} (x e^{x}))

\frac{1}{2 ln ( 2 ) x} + lo g_{4} (e)

Solution steps

\frac{d}{d x} (lo g_{4} (x e^{x}))

Simplify

lo g_{4} (x e^{x}) : lo g_{4} (x) + lo g_{4} (e) x

= \frac{d}{d x} (lo g_{4} (x) + lo g_{4} (e) x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (lo g_{4} (x)) + \frac{d}{d x} (lo g_{4} (e) x)

\frac{d}{d x} (lo g_{4} (x)) = \frac{1}{2 ln ( 2 ) x}

\frac{d}{d x} (lo g_{4} (e) x) = lo g_{4} (e)

= \frac{1}{2 ln ( 2 ) x} + lo g_{4} (e)

y^{^{''}} + 9 y = 4 sec (3 t)

\frac{d y}{d x} + \frac{2 y}{x + 4} = (x + 4)^{7}

\int \frac{x ^{4}}{x ^{2} + 4 x + 4} d x

\int 5 sec^{6} (x) d x

\int \frac{x}{3 x + 1} d x

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