What is the derivative of f(x)=xsqrt(x^2+19) ?

The derivative of f(x)=xsqrt(x^2+19) is (2x^2+19)/(sqrt(x^2+19))

What is the first derivative of f(x)=xsqrt(x^2+19) ?

The first derivative of f(x)=xsqrt(x^2+19) is (2x^2+19)/(sqrt(x^2+19))

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derivative of

f (x) = x x^{2} + 19

\frac{2 x ^{2} + 19}{x ^{2} + 19}

Solution steps

\frac{d}{d x} (x x^{2} + 19)

Apply the Product Rule:

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

f = x, g = x^{2} + 19

= \frac{d x}{d x} x^{2} + 19 + \frac{d}{d x} (x^{2} + 19) x

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

\frac{d}{d x} (x^{2} + 19) = \frac{x}{x ^{2} + 19}

= 1 \cdot x^{2} + 19 + \frac{x}{x ^{2} + 19} x

Simplify

1 \cdot x^{2} + 19 + \frac{x}{x ^{2} + 19} x : \frac{2 x ^{2} + 19}{x ^{2} + 19}

= \frac{2 x ^{2} + 19}{x ^{2} + 19}

f (x) = x

y^{^{'}} x - y = 3

\frac{1}{2}

f (x) = - 2 x

(- 3, - 5), (2, 5)

What is the derivative of f(x)=xsqrt(x^2+19) ?
The derivative of f(x)=xsqrt(x^2+19) is (2x^2+19)/(sqrt(x^2+19))
What is the first derivative of f(x)=xsqrt(x^2+19) ?
The first derivative of f(x)=xsqrt(x^2+19) is (2x^2+19)/(sqrt(x^2+19))