What is the derivative of f(x)=6x^{4/3}-2/3 x^{3/2}+x^2-7x+1 ?

The derivative of f(x)=6x^{4/3}-2/3 x^{3/2}+x^2-7x+1 is 2x+8x^{1/3}-x^{1/2}-7

What is the first derivative of f(x)=6x^{4/3}-2/3 x^{3/2}+x^2-7x+1 ?

The first derivative of f(x)=6x^{4/3}-2/3 x^{3/2}+x^2-7x+1 is 2x+8x^{1/3}-x^{1/2}-7

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derivative of

f (x) = 6 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 7 x + 1

2 x + 8 x^{\frac{1}{3}} - x^{\frac{1}{2}} - 7

Solution steps

\frac{d}{d x} (6 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 7 x + 1)

Simplify

6 x^{\frac{4}{3}} - \frac{2}{3} x^{\frac{3}{2}} + x^{2} - 7 x + 1 : 6 x^{\frac{4}{3}} - \frac{2 x ^{\frac{3}{2}}}{3} + x^{2} - 7 x + 1

= \frac{d}{d x} (6 x^{\frac{4}{3}} - \frac{2 x ^{\frac{3}{2}}}{3} + x^{2} - 7 x + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (6 x^{\frac{4}{3}}) - \frac{d}{d x} (\frac{2 x ^{\frac{3}{2}}}{3}) + \frac{d}{d x} (x^{2}) - \frac{d}{d x} (7 x) + \frac{d}{d x} (1)

\frac{d}{d x} (6 x^{\frac{4}{3}}) = 8 x^{\frac{1}{3}}

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

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

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

\frac{d}{d x} (1) = 0

= 8 x^{\frac{1}{3}} - x^{\frac{1}{2}} + 2 x - 7 + 0

Simplify

= 2 x + 8 x^{\frac{1}{3}} - x^{\frac{1}{2}} - 7

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

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{- t}

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

\int_{0}^{1} 11 x x^{2} + 36 d x

x y^{^{'}} + 2 y^{^{'}} - (x + 2) sin (x) + y = 0

What is the derivative of f(x)=6x^{4/3}-2/3 x^{3/2}+x^2-7x+1 ?
The derivative of f(x)=6x^{4/3}-2/3 x^{3/2}+x^2-7x+1 is 2x+8x^{1/3}-x^{1/2}-7
What is the first derivative of f(x)=6x^{4/3}-2/3 x^{3/2}+x^2-7x+1 ?
The first derivative of f(x)=6x^{4/3}-2/3 x^{3/2}+x^2-7x+1 is 2x+8x^{1/3}-x^{1/2}-7