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

The derivative of 4^xlog_{4}(x) is ln(2)*2^{2x+1}log_{4}(x)+(2^{2x-1})/(ln(2)x)

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

The first derivative of 4^xlog_{4}(x) is ln(2)*2^{2x+1}log_{4}(x)+(2^{2x-1})/(ln(2)x)

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

ln (2) \cdot 2^{2 x + 1} lo g_{4} (x) + \frac{2 ^{2 x - 1}}{ln ( 2 ) x}

Solution steps

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

Apply the Product Rule:

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

f = 4^{x}, g = lo g_{4} (x)

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

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

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

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

\frac{1}{2 ln ( 2 ) x} \cdot 4^{x} = \frac{2 ^{2 x - 1}}{ln ( 2 ) x}

= ln (2) \cdot 2^{2 x + 1} lo g_{4} (x) + \frac{2 ^{2 x - 1}}{ln ( 2 ) x}

F (x) = e^{x^{2}}

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

\int_{0}^{\frac{π}{6}} cos (8 x) d x

\int \frac{1}{2} x^{3} d x

\int_{0}^{4} 1 + x d x

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