What is the derivative of 3xsqrt(4x^2+4) ?

The derivative of 3xsqrt(4x^2+4) is (3(4x^2+2))/(sqrt(x^2+1))

What is the first derivative of 3xsqrt(4x^2+4) ?

The first derivative of 3xsqrt(4x^2+4) is (3(4x^2+2))/(sqrt(x^2+1))

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derivative of

3 x 4 x^{2} + 4

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

Solution steps

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

Take the constant out:

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

= 3 \frac{d}{d x} (x 4 x^{2} + 4)

Apply the Product Rule:

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

f = x, g = 4 x^{2} + 4

= 3 (\frac{d x}{d x} 4 x^{2} + 4 + \frac{d}{d x} (4 x^{2} + 4) x)

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

\frac{d}{d x} (4 x^{2} + 4) = \frac{2 x}{x ^{2} + 1}

= 3 (1 \cdot 4 x^{2} + 4 + \frac{2 x}{x ^{2} + 1} x)

Simplify

3 (1 \cdot 4 x^{2} + 4 + \frac{2 x}{x ^{2} + 1} x) : \frac{3 ( 4 x ^{2} + 2 )}{x ^{2} + 1}

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

\int x 25 + x^{2} d x

\frac{\partial}{\partial x} (sin (z) \cdot cos (x \cdot y))

(2 + x) y^{^{'}} = 5 y

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

\frac{1}{1 - x + \frac{x ^{2}}{2}}, 0

What is the derivative of 3xsqrt(4x^2+4) ?
The derivative of 3xsqrt(4x^2+4) is (3(4x^2+2))/(sqrt(x^2+1))
What is the first derivative of 3xsqrt(4x^2+4) ?
The first derivative of 3xsqrt(4x^2+4) is (3(4x^2+2))/(sqrt(x^2+1))