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Popular >

derivative of 1/(2sqrt(x)+1/(5x^{4/5)})

Solution

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

Solution

- \frac{1}{4 x ^{\frac{3}{2}}} - \frac{4}{25 x ^{\frac{9}{5}}}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (\frac{1}{5 x ^{\frac{4}{5}}}) = - \frac{4}{25 x ^{\frac{9}{5}}}

= - \frac{1}{4 x ^{\frac{3}{2}}} - \frac{4}{25 x ^{\frac{9}{5}}}

Popular Examples

(\partial)/(\partial x)(6xcos(x)sin(y))

\frac{\partial}{\partial x} (6 x cos (x) sin (y))

(\partial)/(\partial z)(4ln((3xy)/(7z)))

\frac{\partial}{\partial z} (4 ln (\frac{3 x y}{7 z}))

y^{''}+289y=sec(17x),y(0)=3,y^'(0)=3

y^{^{''}} + 289 y = sec (17 x), y (0) = 3, y^{^{'}} (0) = 3

tangent of y=e^{4x}cos(pix),(0,1)tangent of

y = e^{4 x} cos (π x), (0, 1)

derivative of e^{0.5x}

\frac{d}{d x} (e^{0.5 x})

Frequently Asked Questions (FAQ)

What is the derivative of 1/(2sqrt(x)+1/(5x^{4/5)}) ?
The derivative of 1/(2sqrt(x)+1/(5x^{4/5)}) is -1/(4x^{3/2)}-4/(25x^{9/5)}
What is the first derivative of 1/(2sqrt(x)+1/(5x^{4/5)}) ?
The first derivative of 1/(2sqrt(x)+1/(5x^{4/5)}) is -1/(4x^{3/2)}-4/(25x^{9/5)}