What is the derivative of ((x^4-5x^3+x^{1/2}))/(x^2) ?

The derivative of ((x^4-5x^3+x^{1/2}))/(x^2) is (4x^5-10x^4-3x^{3/2})/(2x^4)

What is the first derivative of ((x^4-5x^3+x^{1/2}))/(x^2) ?

The first derivative of ((x^4-5x^3+x^{1/2}))/(x^2) is (4x^5-10x^4-3x^{3/2})/(2x^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\frac{( x ^{4} - 5 x ^{3} + x ^{\frac{1}{2}} )}{x ^{2}}

\frac{4 x ^{5} - 10 x ^{4} - 3 x ^{\frac{3}{2}}}{2 x ^{4}}

Solution steps

\frac{d}{d x} (\frac{x ^{4} - 5 x ^{3} + x ^{\frac{1}{2}}}{x ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{4} - 5 x ^{3} + x ^{\frac{1}{2}} ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( x ^{4} - 5 x ^{3} + x ^{\frac{1}{2}} )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (x^{4} - 5 x^{3} + x^{\frac{1}{2}}) = \frac{1 + 8 x ^{\frac{7}{2}} - 30 x ^{\frac{5}{2}}}{2 x ^{\frac{1}{2}}}

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

= \frac{\frac{1 + 8 x ^{\frac{7}{2}} - 30 x ^{\frac{5}{2}}}{2 x ^{\frac{1}{2}}} x ^{2} - 2 x ( x ^{4} - 5 x ^{3} + x ^{\frac{1}{2}} )}{( x ^{2} ) ^{2}}

Simplify

\frac{\frac{1 + 8 x ^{\frac{7}{2}} - 30 x ^{\frac{5}{2}}}{2 x ^{\frac{1}{2}}} x ^{2} - 2 x ( x ^{4} - 5 x ^{3} + x ^{\frac{1}{2}} )}{( x ^{2} ) ^{2}} : \frac{4 x ^{5} - 10 x ^{4} - 3 x ^{\frac{3}{2}}}{2 x ^{4}}

= \frac{4 x ^{5} - 10 x ^{4} - 3 x ^{\frac{3}{2}}}{2 x ^{4}}

\int_{0}^{\infty} 4 x e^{- 2 x} d x

y^{^{''}} + 3 y^{^{'}} + 2 y = 4 t

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

\frac{1}{s ^{2} + 2 \cdot 3 \cdot π \cdot s + π ^{2}}

\int \frac{x}{7 x + 1} d x

What is the derivative of ((x^4-5x^3+x^{1/2}))/(x^2) ?
The derivative of ((x^4-5x^3+x^{1/2}))/(x^2) is (4x^5-10x^4-3x^{3/2})/(2x^4)
What is the first derivative of ((x^4-5x^3+x^{1/2}))/(x^2) ?
The first derivative of ((x^4-5x^3+x^{1/2}))/(x^2) is (4x^5-10x^4-3x^{3/2})/(2x^4)