What is the derivative of (100p)/(0.1p+1) ?

The derivative of (100p)/(0.1p+1) is (100)/((0.1p+1)^2)

What is the first derivative of (100p)/(0.1p+1) ?

The first derivative of (100p)/(0.1p+1) is (100)/((0.1p+1)^2)

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derivative of

\frac{100 p}{0.1 p + 1}

\frac{100}{( 0.1 p + 1 ) ^{2}}

Solution steps

\frac{d}{d p} (\frac{100 p}{0.1 p + 1})

Take the constant out:

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

= 100 \frac{d}{d p} (\frac{p}{0.1 p + 1})

Apply the Quotient Rule:

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

= 100 \cdot \frac{\frac{d p}{d p} ( 0.1 p + 1 ) - \frac{d}{d p} ( 0.1 p + 1 ) p}{( 0.1 p + 1 ) ^{2}}

\frac{d p}{d p} = 1

\frac{d}{d p} (0.1 p + 1) = 0.1

= 100 \cdot \frac{1 \cdot ( 0.1 p + 1 ) - 0.1 p}{( 0.1 p + 1 ) ^{2}}

Simplify

100 \cdot \frac{1 \cdot ( 0.1 p + 1 ) - 0.1 p}{( 0.1 p + 1 ) ^{2}} : \frac{100}{( 0.1 p + 1 ) ^{2}}

= \frac{100}{( 0.1 p + 1 ) ^{2}}

\int 6 x y^{3} d x

\frac{d y}{d x}, 3 x^{3} + 9 x y^{2} = 5 x^{3}

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

\frac{9}{( s - 2 ) ( s - 3 )}

f (x) = (2 x) (4 x^{2} + 6 x^{7} + 8)

What is the derivative of (100p)/(0.1p+1) ?
The derivative of (100p)/(0.1p+1) is (100)/((0.1p+1)^2)
What is the first derivative of (100p)/(0.1p+1) ?
The first derivative of (100p)/(0.1p+1) is (100)/((0.1p+1)^2)