What is the derivative of-4x^{4/3}-2x^{-3/4}+3 ?

The derivative of-4x^{4/3}-2x^{-3/4}+3 is (-32x^{25/12}+9)/(6x^{7/4)}

What is the first derivative of-4x^{4/3}-2x^{-3/4}+3 ?

The first derivative of-4x^{4/3}-2x^{-3/4}+3 is (-32x^{25/12}+9)/(6x^{7/4)}

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

\frac{- 32 x ^{\frac{25}{12}} + 9}{6 x ^{\frac{7}{4}}}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (4 x^{\frac{4}{3}}) = \frac{16 x ^{\frac{1}{3}}}{3}

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

\frac{d}{d x} (3) = 0

= - \frac{16 x ^{\frac{1}{3}}}{3} - (- \frac{3}{2 x ^{\frac{7}{4}}}) + 0

Simplify

- \frac{16 x ^{\frac{1}{3}}}{3} - (- \frac{3}{2 x ^{\frac{7}{4}}}) + 0 : \frac{- 32 x ^{\frac{25}{12}} + 9}{6 x ^{\frac{7}{4}}}

= \frac{- 32 x ^{\frac{25}{12}} + 9}{6 x ^{\frac{7}{4}}}

y = sec (x)

\int - e^{1 - x} d x

\int \frac{2 y}{x} d x

\frac{\partial}{\partial r} (\frac{1}{3} π r^{2} h)

f (x) = \frac{( 5 x - 2 )}{( x ^{2} + 1 )}

What is the derivative of-4x^{4/3}-2x^{-3/4}+3 ?
The derivative of-4x^{4/3}-2x^{-3/4}+3 is (-32x^{25/12}+9)/(6x^{7/4)}
What is the first derivative of-4x^{4/3}-2x^{-3/4}+3 ?
The first derivative of-4x^{4/3}-2x^{-3/4}+3 is (-32x^{25/12}+9)/(6x^{7/4)}