What is the derivative of-(40x)/(9-5x^2) ?

The derivative of-(40x)/(9-5x^2) is -(40(9+5x^2))/((9-5x^2)^2)

What is the first derivative of-(40x)/(9-5x^2) ?

The first derivative of-(40x)/(9-5x^2) is -(40(9+5x^2))/((9-5x^2)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

- \frac{40 x}{9 - 5 x ^{2}}

- \frac{40 ( 9 + 5 x ^{2} )}{( 9 - 5 x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (- \frac{40 x}{9 - 5 x ^{2}})

Take the constant out:

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

= - 40 \frac{d}{d x} (\frac{x}{9 - 5 x ^{2}})

Apply the Quotient Rule:

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

= - 40 \cdot \frac{\frac{d x}{d x} ( 9 - 5 x ^{2} ) - \frac{d}{d x} ( 9 - 5 x ^{2} ) x}{( 9 - 5 x ^{2} ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (9 - 5 x^{2}) = - 10 x

= - 40 \cdot \frac{1 \cdot ( 9 - 5 x ^{2} ) - ( - 10 x ) x}{( 9 - 5 x ^{2} ) ^{2}}

Simplify

- 40 \cdot \frac{1 \cdot ( 9 - 5 x ^{2} ) - ( - 10 x ) x}{( 9 - 5 x ^{2} ) ^{2}} : - \frac{40 ( 9 + 5 x ^{2} )}{( 9 - 5 x ^{2} ) ^{2}}

= - \frac{40 ( 9 + 5 x ^{2} )}{( 9 - 5 x ^{2} ) ^{2}}

\frac{d}{d x} (sin (x) ln (8 x))

\frac{d}{d x} (x^{2} + y z)

y = 8 x^{2} - x^{3} (2.24)

n = 0 \sum \infty \frac{( 8 ^{n} )}{n !}

y^{^{'}} + y = y^{2}

What is the derivative of-(40x)/(9-5x^2) ?
The derivative of-(40x)/(9-5x^2) is -(40(9+5x^2))/((9-5x^2)^2)
What is the first derivative of-(40x)/(9-5x^2) ?
The first derivative of-(40x)/(9-5x^2) is -(40(9+5x^2))/((9-5x^2)^2)