What is the derivative of (log_{6}(x))/(e^{4x)} ?

The derivative of (log_{6}(x))/(e^{4x)} is (1-4xln(x))/(ln(6)e^{4x)x}

What is the first derivative of (log_{6}(x))/(e^{4x)} ?

The first derivative of (log_{6}(x))/(e^{4x)} is (1-4xln(x))/(ln(6)e^{4x)x}

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derivative of

\frac{lo g _{6} ( x )}{e ^{4 x}}

\frac{1 - 4 x ln ( x )}{ln ( 6 ) e ^{4 x} x}

Solution steps

\frac{d}{d x} (\frac{lo g _{6} ( x )}{e ^{4 x}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( lo g _{6} ( x ) ) e ^{4 x} - \frac{d}{d x} ( e ^{4 x} ) lo g _{6} ( x )}{( e ^{4 x} ) ^{2}}

\frac{d}{d x} (lo g_{6} (x)) = \frac{1}{x ln ( 6 )}

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

= \frac{\frac{1}{x l n ( 6 )} e ^{4 x} - e ^{4 x} \cdot 4 lo g _{6} ( x )}{( e ^{4 x} ) ^{2}}

Simplify

\frac{\frac{1}{x l n ( 6 )} e ^{4 x} - e ^{4 x} \cdot 4 lo g _{6} ( x )}{( e ^{4 x} ) ^{2}} : \frac{1 - 4 x ln ( x )}{ln ( 6 ) e ^{4 x} x}

= \frac{1 - 4 x ln ( x )}{ln ( 6 ) e ^{4 x} x}

f (t) = 8 t^{2} - t

4 x, at x = 4

3 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0

y = x, at x = 1

\int_{7}^{e^{5}} \frac{1}{x} d x

What is the derivative of (log_{6}(x))/(e^{4x)} ?
The derivative of (log_{6}(x))/(e^{4x)} is (1-4xln(x))/(ln(6)e^{4x)x}
What is the first derivative of (log_{6}(x))/(e^{4x)} ?
The first derivative of (log_{6}(x))/(e^{4x)} is (1-4xln(x))/(ln(6)e^{4x)x}