What is the derivative of f(x)= 8/(sqrt(x))-4/x ?

The derivative of f(x)= 8/(sqrt(x))-4/x is (-4x^2+4x^{3/2})/(x^{7/2)}

What is the first derivative of f(x)= 8/(sqrt(x))-4/x ?

The first derivative of f(x)= 8/(sqrt(x))-4/x is (-4x^2+4x^{3/2})/(x^{7/2)}

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derivative of

f (x) = \frac{8}{x} - \frac{4}{x}

\frac{- 4 x ^{2} + 4 x ^{\frac{3}{2}}}{x ^{\frac{7}{2}}}

Solution steps

\frac{d}{d x} (\frac{8}{x} - \frac{4}{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{8}{x}) - \frac{d}{d x} (\frac{4}{x})

\frac{d}{d x} (\frac{8}{x}) = - \frac{4}{x ^{\frac{3}{2}}}

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

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

Simplify

- \frac{4}{x ^{\frac{3}{2}}} - (- \frac{4}{x ^{2}}) : \frac{- 4 x ^{2} + 4 x ^{\frac{3}{2}}}{x ^{\frac{7}{2}}}

= \frac{- 4 x ^{2} + 4 x ^{\frac{3}{2}}}{x ^{\frac{7}{2}}}

\int cos^{3} (\frac{x}{6}) d x

y^{^{''}} + 7 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int (\frac{9}{x} - \frac{2}{x ^{3}}) d x

\frac{d}{d x} (x e^{y^{2}})

\frac{d}{d x} (\frac{- x ^{2}}{2})

What is the derivative of f(x)= 8/(sqrt(x))-4/x ?
The derivative of f(x)= 8/(sqrt(x))-4/x is (-4x^2+4x^{3/2})/(x^{7/2)}
What is the first derivative of f(x)= 8/(sqrt(x))-4/x ?
The first derivative of f(x)= 8/(sqrt(x))-4/x is (-4x^2+4x^{3/2})/(x^{7/2)}