What is the derivative of x^8-x^{1/2} ?

The derivative of x^8-x^{1/2} is (-1+16x^{15/2})/(2x^{1/2)}

What is the first derivative of x^8-x^{1/2} ?

The first derivative of x^8-x^{1/2} is (-1+16x^{15/2})/(2x^{1/2)}

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derivative of

x^{8} - x^{\frac{1}{2}}

\frac{- 1 + 16 x ^{\frac{15}{2}}}{2 x ^{\frac{1}{2}}}

Solution steps

\frac{d}{d x} (x^{8} - x^{\frac{1}{2}})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{8}) - \frac{d}{d x} (x^{\frac{1}{2}})

\frac{d}{d x} (x^{8}) = 8 x^{7}

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

= 8 x^{7} - \frac{1}{2 x ^{\frac{1}{2}}}

Simplify

8 x^{7} - \frac{1}{2 x ^{\frac{1}{2}}} : \frac{- 1 + 16 x ^{\frac{15}{2}}}{2 x ^{\frac{1}{2}}}

= \frac{- 1 + 16 x ^{\frac{15}{2}}}{2 x ^{\frac{1}{2}}}

\int \frac{10 x ^{2} - 9 x + 16}{x ^{3} + 4 x} d x

- 3 y^{^{'}} + \frac{3 y}{x} = x^{3} y^{4}

\frac{d y}{d t} = 2 y + y^{3}, y (0) = 1

x^{\frac{2}{7}}

4 x - 3

What is the derivative of x^8-x^{1/2} ?
The derivative of x^8-x^{1/2} is (-1+16x^{15/2})/(2x^{1/2)}
What is the first derivative of x^8-x^{1/2} ?
The first derivative of x^8-x^{1/2} is (-1+16x^{15/2})/(2x^{1/2)}