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

The derivative of f(x)=2xsqrt(2x^2+4) is (2(4x^2+4))/(sqrt(2x^2+4))

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

The first derivative of f(x)=2xsqrt(2x^2+4) is (2(4x^2+4))/(sqrt(2x^2+4))

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derivative of

f (x) = 2 x 2 x^{2} + 4

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

Solution steps

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

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

y^{^{'}} = (3 t + 1)^{2}

f (x) = \frac{2 x}{- x + 2}

\frac{4 d u}{d t} = u^{2}

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