What is the derivative of f(x)=x(3x^4-sqrt(x)) ?

The derivative of f(x)=x(3x^4-sqrt(x)) is (-3sqrt(x)+30x^4)/2

What is the first derivative of f(x)=x(3x^4-sqrt(x)) ?

The first derivative of f(x)=x(3x^4-sqrt(x)) is (-3sqrt(x)+30x^4)/2

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derivative of

f (x) = x (3 x^{4} - x)

\frac{- 3 x + 30 x ^{4}}{2}

Solution steps

\frac{d}{d x} (x (3 x^{4} - x))

Apply the Product Rule:

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

f = x, g = 3 x^{4} - x

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

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

\frac{d}{d x} (3 x^{4} - x) = 12 x^{3} - \frac{1}{2 x}

= 1 \cdot (3 x^{4} - x) + (12 x^{3} - \frac{1}{2 x}) x

Simplify

1 \cdot (3 x^{4} - x) + (12 x^{3} - \frac{1}{2 x}) x : \frac{- 3 x + 30 x ^{4}}{2}

= \frac{- 3 x + 30 x ^{4}}{2}

10 lo g_{10} (x)

x \to 9 lim (\frac{6}{9})

\frac{\partial}{\partial x} (x^{2} + 4 y^{2} + 3 z^{2})

y = 4 x^{2}, (- 1, 4)

x y^{^{'}} - 2 y = 2 x^{2} + 2 x + 1

What is the derivative of f(x)=x(3x^4-sqrt(x)) ?
The derivative of f(x)=x(3x^4-sqrt(x)) is (-3sqrt(x)+30x^4)/2
What is the first derivative of f(x)=x(3x^4-sqrt(x)) ?
The first derivative of f(x)=x(3x^4-sqrt(x)) is (-3sqrt(x)+30x^4)/2