What is the derivative of 3x^2+2yx ?

The derivative of 3x^2+2yx is 6x+2(x(dy)/(dx)+y)

What is the first derivative of 3x^2+2yx ?

The first derivative of 3x^2+2yx is 6x+2(x(dy)/(dx)+y)

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 6 x + 2 (x \frac{d y}{d x} + y)

- 4 x^{2} + 6.25 x + 8, 2.5 x + 7, 0

y = 4, 2 x, \frac{( - 3 x + 7 )}{2}

n = 1 \sum \infty (1 + \frac{k}{n})^{n}

x \to - 8 - lim (\frac{- 5 x + 2}{x + 8})

\int 3 x sin (3 x) d x

What is the derivative of 3x^2+2yx ?
The derivative of 3x^2+2yx is 6x+2(x(dy)/(dx)+y)
What is the first derivative of 3x^2+2yx ?
The first derivative of 3x^2+2yx is 6x+2(x(dy)/(dx)+y)