What is the derivative of y=(200)/(0.9+0.1e^{2x)} ?

The derivative of y=(200)/(0.9+0.1e^{2x)} is -(40e^{2x})/((0.9+0.1e^{2x))^2}

What is the first derivative of y=(200)/(0.9+0.1e^{2x)} ?

The first derivative of y=(200)/(0.9+0.1e^{2x)} is -(40e^{2x})/((0.9+0.1e^{2x))^2}

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derivative of

y = \frac{200}{0.9 + 0.1 e ^{2 x}}

- \frac{40 e ^{2 x}}{( 0.9 + 0.1 e ^{2 x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{200}{0.9 + 0.1 e ^{2 x}})

Take the constant out:

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

= 200 \frac{d}{d x} (\frac{1}{0.9 + 0.1 e ^{2 x}})

Apply exponent rule:

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

= 200 \frac{d}{d x} ((0.9 + 0.1 e^{2 x})^{- 1})

Apply the chain rule:

- \frac{1}{( 0.9 + 0.1 e ^{2 x} ) ^{2}} \frac{d}{d x} (0.9 + 0.1 e^{2 x})

= - \frac{1}{( 0.9 + 0.1 e ^{2 x} ) ^{2}} \frac{d}{d x} (0.9 + 0.1 e^{2 x})

\frac{d}{d x} (0.9 + 0.1 e^{2 x}) = 0.2 e^{2 x}

= 200 (- \frac{1}{( 0.9 + 0.1 e ^{2 x} ) ^{2}} \cdot 0.2 e^{2 x})

Simplify

200 (- \frac{1}{( 0.9 + 0.1 e ^{2 x} ) ^{2}} \cdot 0.2 e^{2 x}) : - \frac{40 e ^{2 x}}{( 0.9 + 0.1 e ^{2 x} ) ^{2}}

= - \frac{40 e ^{2 x}}{( 0.9 + 0.1 e ^{2 x} ) ^{2}}

x \to 0 lim (\frac{2}{x ^{2} + x})

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

\frac{\partial}{\partial v} (2 v cos (u))

0.882 A^{0.842}

- 8 t^{2}

What is the derivative of y=(200)/(0.9+0.1e^{2x)} ?
The derivative of y=(200)/(0.9+0.1e^{2x)} is -(40e^{2x})/((0.9+0.1e^{2x))^2}
What is the first derivative of y=(200)/(0.9+0.1e^{2x)} ?
The first derivative of y=(200)/(0.9+0.1e^{2x)} is -(40e^{2x})/((0.9+0.1e^{2x))^2}