What is the derivative of y=(x^3)/(5-x^2) ?

The derivative of y=(x^3)/(5-x^2) is (15x^2-x^4)/((5-x^2)^2)

What is the first derivative of y=(x^3)/(5-x^2) ?

The first derivative of y=(x^3)/(5-x^2) is (15x^2-x^4)/((5-x^2)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = \frac{x ^{3}}{5 - x ^{2}}

\frac{15 x ^{2} - x ^{4}}{( 5 - x ^{2} ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{3 x ^{2} ( 5 - x ^{2} ) - ( - 2 x ) x ^{3}}{( 5 - x ^{2} ) ^{2}} : \frac{15 x ^{2} - x ^{4}}{( 5 - x ^{2} ) ^{2}}

= \frac{15 x ^{2} - x ^{4}}{( 5 - x ^{2} ) ^{2}}

\frac{\partial}{\partial x} (ln (x^{2} + y^{4}))

y = 2 x^{2} + 5 x + 6, (- 2, 0)

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

\int 4 \frac{cos ( x )}{sin ^{2} ( x )} d x

2 x^{2} - 7 x + 5

What is the derivative of y=(x^3)/(5-x^2) ?
The derivative of y=(x^3)/(5-x^2) is (15x^2-x^4)/((5-x^2)^2)
What is the first derivative of y=(x^3)/(5-x^2) ?
The first derivative of y=(x^3)/(5-x^2) is (15x^2-x^4)/((5-x^2)^2)