What is the derivative of f(x)=(0.3(9^x))/(x^3) ?

The derivative of f(x)=(0.3(9^x))/(x^3) is (0.3*9^x(2.19722…x-3))/(x^4)

What is the first derivative of f(x)=(0.3(9^x))/(x^3) ?

The first derivative of f(x)=(0.3(9^x))/(x^3) is (0.3*9^x(2.19722…x-3))/(x^4)

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derivative of

f (x) = \frac{0.3 ( 9 ^{x} )}{x ^{3}}

\frac{0.3 \cdot 9 ^{x} ( 2.19722 \dots x - 3 )}{x ^{4}}

Solution steps

\frac{d}{d x} (\frac{0.3 \cdot 9 ^{x}}{x ^{3}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.3 \frac{d}{d x} (\frac{9 ^{x}}{x ^{3}})

Apply the Quotient Rule:

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

= 0.3 \cdot \frac{\frac{d}{d x} ( 9 ^{x} ) x ^{3} - \frac{d}{d x} ( x ^{3} ) 9 ^{x}}{( x ^{3} ) ^{2}}

\frac{d}{d x} (9^{x}) = 2 ln (3) \cdot 9^{x}

\frac{d}{d x} (x^{3}) = 3 x^{2}

= 0.3 \cdot \frac{2 ln ( 3 ) \cdot 9 ^{x} x ^{3} - 3 x ^{2} \cdot 9 ^{x}}{( x ^{3} ) ^{2}}

Simplify

0.3 \cdot \frac{2 ln ( 3 ) \cdot 9 ^{x} x ^{3} - 3 x ^{2} \cdot 9 ^{x}}{( x ^{3} ) ^{2}} : \frac{0.3 \cdot 9 ^{x} ( 2.19722 \dots x - 3 )}{x ^{4}}

= \frac{0.3 \cdot 9 ^{x} ( 2.19722 \dots x - 3 )}{x ^{4}}

y^{^{''}} - 4 y^{^{'}} + 4 y = (12 x^{2} - 6 x) e^{2} x

n = 1 \sum \infty \frac{8 ^{n}}{n !}

\int \frac{x}{3 x + 2} d x

8 x - x^{2}

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

What is the derivative of f(x)=(0.3(9^x))/(x^3) ?
The derivative of f(x)=(0.3(9^x))/(x^3) is (0.3*9^x(2.19722…x-3))/(x^4)
What is the first derivative of f(x)=(0.3(9^x))/(x^3) ?
The first derivative of f(x)=(0.3(9^x))/(x^3) is (0.3*9^x(2.19722…x-3))/(x^4)