What is the d/(d{x)}(4{x}-3{x}^2-{y}^2-({z}-1)^2) ?

The d/(d{x)}(4{x}-3{x}^2-{y}^2-({z}-1)^2) is -6{x}+4

What is the first d/(d{x)}(4{x}-3{x}^2-{y}^2-({z}-1)^2) ?

The first d/(d{x)}(4{x}-3{x}^2-{y}^2-({z}-1)^2) is -6{x}+4

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\frac{d}{d x} (4 x - 3 x^{2} - y^{2} - (z - 1)^{2})

- 6 x + 4

Solution steps

\frac{d}{d x} (4 x - 3 x^{2} - y^{2} - (z - 1)^{2})

Simplify

4 x - 3 x^{2} - y^{2} - (z - 1)^{2} : 4 x - 3 x^{2} - y^{2} - z^{2} + 2 z - 1

= \frac{d}{d x} (4 x - 3 x^{2} - y^{2} - z^{2} + 2 z - 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (4 x) - \frac{d}{d x} (3 x^{2}) - \frac{d}{d x} (y^{2}) - \frac{d}{d x} (z^{2}) + \frac{d}{d x} (2 z) - \frac{d}{d x} (1)

\frac{d}{d x} (4 x) = 4

\frac{d}{d x} (3 x^{2}) = 6 x

\frac{d}{d x} (y^{2}) = 0

\frac{d}{d x} (z^{2}) = 0

\frac{d}{d x} (2 z) = 0

\frac{d}{d x} (1) = 0

= 4 - 6 x - 0 - 0 + 0 - 0

Simplify

= - 6 x + 4

\frac{3}{x} - e^{4}

\frac{d}{d x} (- 5 x (2 x - 3)^{4})

\frac{\partial}{\partial v} (u (v, w) + v \cdot e^{w})

\int sin (2 x) e^{x} d x

p (x) = x^{4} - 3 x^{3} + 10 x^{2} - 6 x - 20

What is the d/(d{x)}(4{x}-3{x}^2-{y}^2-({z}-1)^2) ?
The d/(d{x)}(4{x}-3{x}^2-{y}^2-({z}-1)^2) is -6{x}+4
What is the first d/(d{x)}(4{x}-3{x}^2-{y}^2-({z}-1)^2) ?
The first d/(d{x)}(4{x}-3{x}^2-{y}^2-({z}-1)^2) is -6{x}+4