What is the derivative of 2yx ?

The derivative of 2yx is 2(x(dy)/(dx)+y)

What is the first derivative of 2yx ?

The first derivative of 2yx is 2(x(dy)/(dx)+y)

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

2 (x \frac{d y}{d x} + y)

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y, g = x

= 2 (\frac{d y}{d x} x + \frac{d x}{d x} y)

\frac{d x}{d x} = 1

= 2 (\frac{d y}{d x} x + 1 \cdot y)

Simplify

= 2 (x \frac{d y}{d x} + y)

\frac{\partial}{\partial x} (4 x^{2} y^{3} + 2)

\frac{1}{s ( s ^{2} + 7 s + 12 )}

\int sin^{2} (x) \cdot cos^{3} (x) d x

x^{2}

\frac{d}{d x} (\frac{5 x - 1}{( 6 x ^{2} - 2 ) ^{2}})

What is the derivative of 2yx ?
The derivative of 2yx is 2(x(dy)/(dx)+y)
What is the first derivative of 2yx ?
The first derivative of 2yx is 2(x(dy)/(dx)+y)