What is the d/(d{x)}(3({x}^2)i+5({y}^3)j+{z}k) ?

The d/(d{x)}(3({x}^2)i+5({y}^3)j+{z}k) is 6i{x}

What is the first d/(d{x)}(3({x}^2)i+5({y}^3)j+{z}k) ?

The first d/(d{x)}(3({x}^2)i+5({y}^3)j+{z}k) is 6i{x}

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\frac{d}{d x} (3 (x^{2}) i + 5 (y^{3}) j + z k)

6 i x

Solution steps

\frac{d}{d x} (3 (x^{2}) i + 5 (y^{3}) j + z k)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (3 x^{2} i) + \frac{d}{d x} (5 y^{3} j) + \frac{d}{d x} (z k)

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

\frac{d}{d x} (5 y^{3} j) = 0

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

= 6 i x + 0 + 0

Simplify

= 6 i x

x \to - 5 + lim (3 [x] + 4)

\frac{\partial}{\partial z} (5 z e^{x yz})

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\int_{0}^{4} \frac{1}{16 + x ^{2}} d x

What is the d/(d{x)}(3({x}^2)i+5({y}^3)j+{z}k) ?
The d/(d{x)}(3({x}^2)i+5({y}^3)j+{z}k) is 6i{x}
What is the first d/(d{x)}(3({x}^2)i+5({y}^3)j+{z}k) ?
The first d/(d{x)}(3({x}^2)i+5({y}^3)j+{z}k) is 6i{x}