What is the derivative of (0.09(5^x))/(x^3) ?

The derivative of (0.09(5^x))/(x^3) is (0.09*5^x(1.60943…x-3))/(x^4)

What is the first derivative of (0.09(5^x))/(x^3) ?

The first derivative of (0.09(5^x))/(x^3) is (0.09*5^x(1.60943…x-3))/(x^4)

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derivative of

\frac{0.09 ( 5 ^{x} )}{x ^{3}}

\frac{0.09 \cdot 5 ^{x} ( 1.60943 \dots x - 3 )}{x ^{4}}

Solution steps

\frac{d}{d x} (\frac{0.09 \cdot 5 ^{x}}{x ^{3}})

Take the constant out:

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

= 0.09 \frac{d}{d x} (\frac{5 ^{x}}{x ^{3}})

Apply the Quotient Rule:

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

= 0.09 \cdot \frac{\frac{d}{d x} ( 5 ^{x} ) x ^{3} - \frac{d}{d x} ( x ^{3} ) 5 ^{x}}{( x ^{3} ) ^{2}}

\frac{d}{d x} (5^{x}) = 5^{x} ln (5)

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

= 0.09 \cdot \frac{5 ^{x} ln ( 5 ) x ^{3} - 3 x ^{2} \cdot 5 ^{x}}{( x ^{3} ) ^{2}}

Simplify

0.09 \cdot \frac{5 ^{x} ln ( 5 ) x ^{3} - 3 x ^{2} \cdot 5 ^{x}}{( x ^{3} ) ^{2}} : \frac{0.09 \cdot 5 ^{x} ( 1.60943 \dots x - 3 )}{x ^{4}}

= \frac{0.09 \cdot 5 ^{x} ( 1.60943 \dots x - 3 )}{x ^{4}}

\frac{8 x}{7 - cot ( x )}

\frac{\partial}{\partial y} (3 ln (x) + ln (y))

y^{^{''}} + 16 y^{^{'}} = 0, y (0) = 2, y (0) = - 2

x \to 1 lim (3 x + 7)

y = 4 x + 84, (4, 10)

What is the derivative of (0.09(5^x))/(x^3) ?
The derivative of (0.09(5^x))/(x^3) is (0.09*5^x(1.60943…x-3))/(x^4)
What is the first derivative of (0.09(5^x))/(x^3) ?
The first derivative of (0.09(5^x))/(x^3) is (0.09*5^x(1.60943…x-3))/(x^4)