What is the derivative of y= 2/(1-9x^2) ?

The derivative of y= 2/(1-9x^2) is (36x)/((1-9x^2)^2)

What is the first derivative of y= 2/(1-9x^2) ?

The first derivative of y= 2/(1-9x^2) is (36x)/((1-9x^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{2}{1 - 9 x ^{2}}

\frac{36 x}{( 1 - 9 x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2}{1 - 9 x ^{2}})

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{1}{1 - 9 x ^{2}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 2 \frac{d}{d x} ((1 - 9 x^{2})^{- 1})

Apply the chain rule:

- \frac{1}{( 1 - 9 x ^{2} ) ^{2}} \frac{d}{d x} (1 - 9 x^{2})

= - \frac{1}{( 1 - 9 x ^{2} ) ^{2}} \frac{d}{d x} (1 - 9 x^{2})

\frac{d}{d x} (1 - 9 x^{2}) = - 18 x

= 2 (- \frac{1}{( 1 - 9 x ^{2} ) ^{2}} (- 18 x))

Simplify

2 (- \frac{1}{( 1 - 9 x ^{2} ) ^{2}} (- 18 x)) : \frac{36 x}{( 1 - 9 x ^{2} ) ^{2}}

= \frac{36 x}{( 1 - 9 x ^{2} ) ^{2}}

\int_{0}^{\frac{π}{2}} x cos (\frac{x}{3}) d x

\frac{2 ln ( x )}{x}

\frac{d y}{d x} = 2 y (9 - y)

\frac{d x}{d t} = x (1 - \frac{1}{2500} x)

\int (x + 4) 8 x + x^{2} d x

What is the derivative of y= 2/(1-9x^2) ?
The derivative of y= 2/(1-9x^2) is (36x)/((1-9x^2)^2)
What is the first derivative of y= 2/(1-9x^2) ?
The first derivative of y= 2/(1-9x^2) is (36x)/((1-9x^2)^2)