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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

f (x) = \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}}

(100 - x^{2}) \frac{d y}{d x} = 7 y

6 sin (x) - 6 cos (2 x), [0, \frac{π}{2}]

\frac{\partial}{\partial x} (- x sin (x y))

\frac{d}{d x} (x \cdot e^{x} \cdot sin (x))

\int_{8}^{10} \frac{y}{y ^{2} - 4 y - 5} d y

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