What is the derivative of 5x^4-3x^2+1/(sqrt(x)) ?

The derivative of 5x^4-3x^2+1/(sqrt(x)) is (-1+40x^{9/2}-12x^{5/2})/(2x^{3/2)}

What is the first derivative of 5x^4-3x^2+1/(sqrt(x)) ?

The first derivative of 5x^4-3x^2+1/(sqrt(x)) is (-1+40x^{9/2}-12x^{5/2})/(2x^{3/2)}

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\frac{d}{d x} (5 x^{4} - 3 x^{2} + \frac{1}{x})

\frac{- 1 + 40 x ^{\frac{9}{2}} - 12 x ^{\frac{5}{2}}}{2 x ^{\frac{3}{2}}}

Solution steps

\frac{d}{d x} (5 x^{4} - 3 x^{2} + \frac{1}{x})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (5 x^{4}) - \frac{d}{d x} (3 x^{2}) + \frac{d}{d x} (\frac{1}{x})

\frac{d}{d x} (5 x^{4}) = 20 x^{3}

\frac{d}{d x} (3 x^{2}) = 6 x

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

= 20 x^{3} - 6 x - \frac{1}{2 x ^{\frac{3}{2}}}

Simplify

20 x^{3} - 6 x - \frac{1}{2 x ^{\frac{3}{2}}} : \frac{- 1 + 40 x ^{\frac{9}{2}} - 12 x ^{\frac{5}{2}}}{2 x ^{\frac{3}{2}}}

= \frac{- 1 + 40 x ^{\frac{9}{2}} - 12 x ^{\frac{5}{2}}}{2 x ^{\frac{3}{2}}}

x \to 4 lim (∣ x + 4 ∣)

x^{2} y^{^{'}} + x y = 4

\frac{\partial}{\partial x} (f (x) (x f (x) (x f (x))))

\int e^{2 x} arcsin (e^{x}) d x

(x^{3} + x)^{6}

What is the derivative of 5x^4-3x^2+1/(sqrt(x)) ?
The derivative of 5x^4-3x^2+1/(sqrt(x)) is (-1+40x^{9/2}-12x^{5/2})/(2x^{3/2)}
What is the first derivative of 5x^4-3x^2+1/(sqrt(x)) ?
The first derivative of 5x^4-3x^2+1/(sqrt(x)) is (-1+40x^{9/2}-12x^{5/2})/(2x^{3/2)}