What is the derivative of 2/((1+a*e^{b*x^2)}) ?

The derivative of 2/((1+a*e^{b*x^2)}) is -(4abe^{bx^2}x)/((1+ae^{bx^2))^2}

What is the first derivative of 2/((1+a*e^{b*x^2)}) ?

The first derivative of 2/((1+a*e^{b*x^2)}) is -(4abe^{bx^2}x)/((1+ae^{bx^2))^2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

\frac{d}{d x} (\frac{2}{( 1 + a \cdot e ^{b \cdot x^{2}} )})

- \frac{4 ab e ^{b x^{2}} x}{( 1 + a e ^{b x^{2}} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2}{( 1 + a e ^{b x^{2}} )})

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{1}{1 + a e ^{b x^{2}}})

Apply exponent rule:

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

= 2 \frac{d}{d x} ((1 + a e^{b x^{2}})^{- 1})

Apply the chain rule:

- \frac{1}{( 1 + a e ^{b x^{2}} ) ^{2}} \frac{d}{d x} (1 + a e^{b x^{2}})

= - \frac{1}{( 1 + a e ^{b x^{2}} ) ^{2}} \frac{d}{d x} (1 + a e^{b x^{2}})

\frac{d}{d x} (1 + a e^{b x^{2}}) = 2 ab e^{b x^{2}} x

= 2 (- \frac{1}{( 1 + a e ^{b x^{2}} ) ^{2}} \cdot 2 ab e^{b x^{2}} x)

Simplify

2 (- \frac{1}{( 1 + a e ^{b x^{2}} ) ^{2}} \cdot 2 ab e^{b x^{2}} x) : - \frac{4 ab e ^{b x^{2}} x}{( 1 + a e ^{b x^{2}} ) ^{2}}

= - \frac{4 ab e ^{b x^{2}} x}{( 1 + a e ^{b x^{2}} ) ^{2}}

What is the derivative of 2/((1+a*e^{b*x^2)}) ?
The derivative of 2/((1+a*e^{b*x^2)}) is -(4abe^{bx^2}x)/((1+ae^{bx^2))^2}
What is the first derivative of 2/((1+a*e^{b*x^2)}) ?
The first derivative of 2/((1+a*e^{b*x^2)}) is -(4abe^{bx^2}x)/((1+ae^{bx^2))^2}