What is the derivative of f(x)=xsqrt(2x^2+7) ?

The derivative of f(x)=xsqrt(2x^2+7) is (4x^2+7)/(sqrt(2x^2+7))

What is the first derivative of f(x)=xsqrt(2x^2+7) ?

The first derivative of f(x)=xsqrt(2x^2+7) is (4x^2+7)/(sqrt(2x^2+7))

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derivative of

f (x) = x 2 x^{2} + 7

\frac{4 x ^{2} + 7}{2 x ^{2} + 7}

Solution steps

\frac{d}{d x} (x 2 x^{2} + 7)

Apply the Product Rule:

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

f = x, g = 2 x^{2} + 7

= \frac{d x}{d x} 2 x^{2} + 7 + \frac{d}{d x} (2 x^{2} + 7) x

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

\frac{d}{d x} (2 x^{2} + 7) = \frac{2 x}{2 x ^{2} + 7}

= 1 \cdot 2 x^{2} + 7 + \frac{2 x}{2 x ^{2} + 7} x

Simplify

1 \cdot 2 x^{2} + 7 + \frac{2 x}{2 x ^{2} + 7} x : \frac{4 x ^{2} + 7}{2 x ^{2} + 7}

= \frac{4 x ^{2} + 7}{2 x ^{2} + 7}

\frac{ln ( x ) - 1}{ln ^{2} ( x )}

\int sin^{2} (5 θ) d θ

\int \frac{x ^{2} - 5 x + 3}{x - 7} d x

x^{2} + y^{2} = x y y^{^{'}}

e^{- 2 s} (\frac{1}{4 s})

What is the derivative of f(x)=xsqrt(2x^2+7) ?
The derivative of f(x)=xsqrt(2x^2+7) is (4x^2+7)/(sqrt(2x^2+7))
What is the first derivative of f(x)=xsqrt(2x^2+7) ?
The first derivative of f(x)=xsqrt(2x^2+7) is (4x^2+7)/(sqrt(2x^2+7))