What is the derivative of 1/(1+|x|+1/(1+|x-3|)) ?

The derivative of 1/(1+|x|+1/(1+|x-3|)) is

What is the first derivative of 1/(1+|x|+1/(1+|x-3|)) ?

The first derivative of 1/(1+|x|+1/(1+|x-3|)) is

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\frac{d}{d x} (\frac{1}{1 + ∣ x ∣} + \frac{1}{1 + ∣ x - 3 ∣})

\frac{- 4 x ^{3} - x ^{3} ∣ x ∣ + 18 x ^{2} + 3 x ^{2} ∣ x ∣ - x ∣ x - 3 ∣ ^{3} - 18 x - x ∣ x - 3 ∣ - x ∣ x ∣ + 3 ∣ x ∣}{∣ x ∣ ∣ x - 3 ∣ ( ∣ x ∣ + 1 ) ^{2} ( ∣ x - 3 ∣ + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{1 + ∣ x ∣} + \frac{1}{1 + ∣ x - 3 ∣})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{1}{1 + ∣ x ∣}) + \frac{d}{d x} (\frac{1}{1 + ∣ x - 3 ∣})

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

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

= - \frac{x}{∣ x ∣ ( 1 + ∣ x ∣ ) ^{2}} - \frac{x - 3}{∣ x - 3 ∣ ( 1 + ∣ x - 3 ∣ ) ^{2}}

Simplify

- \frac{x}{∣ x ∣ ( 1 + ∣ x ∣ ) ^{2}} - \frac{x - 3}{∣ x - 3 ∣ ( 1 + ∣ x - 3 ∣ ) ^{2}} : \frac{- 4 x ^{3} - x ^{3} ∣ x ∣ + 18 x ^{2} + 3 x ^{2} ∣ x ∣ - x ∣ x - 3 ∣ ^{3} - 18 x - x ∣ x - 3 ∣ - x ∣ x ∣ + 3 ∣ x ∣}{∣ x ∣ ∣ x - 3 ∣ ( ∣ x ∣ + 1 ) ^{2} ( ∣ x - 3 ∣ + 1 ) ^{2}}

= \frac{- 4 x ^{3} - x ^{3} ∣ x ∣ + 18 x ^{2} + 3 x ^{2} ∣ x ∣ - x ∣ x - 3 ∣ ^{3} - 18 x - x ∣ x - 3 ∣ - x ∣ x ∣ + 3 ∣ x ∣}{∣ x ∣ ∣ x - 3 ∣ ( ∣ x ∣ + 1 ) ^{2} ( ∣ x - 3 ∣ + 1 ) ^{2}}

\frac{\partial}{\partial y} (2 x - 3 yz)

\frac{d}{d r} (\frac{200 r ( r + 2 )}{( 2 + 3 r ) ^{2}})

y = 9 - 10 x, at x = - 1

\frac{d}{d x} (2 x y^{3})

\frac{\partial}{\partial y} (e^{s i n (x + y)})

What is the derivative of 1/(1+|x|+1/(1+|x-3|)) ?
The derivative of 1/(1+|x|+1/(1+|x-3|)) is
What is the first derivative of 1/(1+|x|+1/(1+|x-3|)) ?
The first derivative of 1/(1+|x|+1/(1+|x-3|)) is