What is the derivative of (80/(1+0.32x)) ?

The derivative of (80/(1+0.32x)) is -(25.6)/((1+0.32x)^2)

What is the first derivative of (80/(1+0.32x)) ?

The first derivative of (80/(1+0.32x)) is -(25.6)/((1+0.32x)^2)

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\frac{d}{d x} (\frac{80}{1 + 0.32 x})

- \frac{25.6}{( 1 + 0.32 x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{80}{1 + 0.32 x})

Take the constant out:

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

= 80 \frac{d}{d x} (\frac{1}{1 + 0.32 x})

Apply exponent rule:

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

= 80 \frac{d}{d x} ((1 + 0.32 x)^{- 1})

Apply the chain rule:

- \frac{1}{( 1 + 0.32 x ) ^{2}} \frac{d}{d x} (1 + 0.32 x)

= - \frac{1}{( 1 + 0.32 x ) ^{2}} \frac{d}{d x} (1 + 0.32 x)

\frac{d}{d x} (1 + 0.32 x) = 0.32

= 80 (- \frac{1}{( 1 + 0.32 x ) ^{2}} \cdot 0.32)

Simplify

80 (- \frac{1}{( 1 + 0.32 x ) ^{2}} \cdot 0.32) : - \frac{25.6}{( 1 + 0.32 x ) ^{2}}

= - \frac{25.6}{( 1 + 0.32 x ) ^{2}}

x \to 1 lim (e^{x - 1})

\frac{d}{d x} (- \frac{2}{2 x + 5})

f (x) = ln (4 x^{2} + 1)

x \to 1 - lim (\frac{2 x}{x ^{2} - 1})

\frac{\partial}{\partial y} (x^{2} y + x + 1)

What is the derivative of (80/(1+0.32x)) ?
The derivative of (80/(1+0.32x)) is -(25.6)/((1+0.32x)^2)
What is the first derivative of (80/(1+0.32x)) ?
The first derivative of (80/(1+0.32x)) is -(25.6)/((1+0.32x)^2)