What is the derivative of (x^{3.8}/(6^x)) ?

The derivative of (x^{3.8}/(6^x)) is (x^{2.8}(3.8-1.79175…x))/(6^x)

What is the first derivative of (x^{3.8}/(6^x)) ?

The first derivative of (x^{3.8}/(6^x)) is (x^{2.8}(3.8-1.79175…x))/(6^x)

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\frac{d}{d x} (\frac{x ^{3.8}}{6 ^{x}})

\frac{x ^{2.8} ( 3.8 - 1.79175 \dots x )}{6 ^{x}}

Solution steps

\frac{d}{d x} (\frac{x ^{3.8}}{6 ^{x}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{3.8} ) 6 ^{x} - \frac{d}{d x} ( 6 ^{x} ) x ^{3.8}}{( 6 ^{x} ) ^{2}}

\frac{d}{d x} (x^{3.8}) = 3.8 x^{2.8}

\frac{d}{d x} (6^{x}) = 6^{x} ln (6)

= \frac{3.8 x ^{2.8} \cdot 6 ^{x} - 6 ^{x} ln ( 6 ) x ^{3.8}}{( 6 ^{x} ) ^{2}}

Simplify

\frac{3.8 x ^{2.8} \cdot 6 ^{x} - 6 ^{x} ln ( 6 ) x ^{3.8}}{( 6 ^{x} ) ^{2}} : \frac{x ^{2.8} ( 3.8 - 1.79175 \dots x )}{6 ^{x}}

= \frac{x ^{2.8} ( 3.8 - 1.79175 \dots x )}{6 ^{x}}

tan^{2} (6 θ)

\frac{1}{s ^{3} ( s - 1 )}

\frac{\partial}{\partial x} (cos (2 y - 2 x))

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{4} d x

x \to 0 lim (\frac{1}{x} - \frac{1}{∣ x ∣})

What is the derivative of (x^{3.8}/(6^x)) ?
The derivative of (x^{3.8}/(6^x)) is (x^{2.8}(3.8-1.79175…x))/(6^x)
What is the first derivative of (x^{3.8}/(6^x)) ?
The first derivative of (x^{3.8}/(6^x)) is (x^{2.8}(3.8-1.79175…x))/(6^x)