What is the derivative of 9/((x-6^{4/3)}) ?

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

What is the first derivative of 9/((x-6^{4/3)}) ?

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

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\frac{d}{d x} (\frac{9}{( x - 6 ) ^{\frac{4}{3}}})

- \frac{12}{( x - 6 ) ^{\frac{7}{3}}}

Solution steps

\frac{d}{d x} (\frac{9}{( x - 6 ) ^{\frac{4}{3}}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 9 \frac{d}{d x} (\frac{1}{( x - 6 ) ^{\frac{4}{3}}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 9 \frac{d}{d x} ((x - 6)^{- \frac{4}{3}})

Apply the chain rule:

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

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

\frac{d}{d x} (x - 6) = 1

= 9 (- \frac{4}{3 ( x - 6 ) ^{\frac{7}{3}}} \cdot 1)

Simplify

9 (- \frac{4}{3 ( x - 6 ) ^{\frac{7}{3}}} \cdot 1) : - \frac{12}{( x - 6 ) ^{\frac{7}{3}}}

= - \frac{12}{( x - 6 ) ^{\frac{7}{3}}}

\int 10 5 x^{2} - 1 d x

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{2 x} (3 - 5 x)

\int \frac{x + 4}{2 x ^{2} + x - 1} d x

t an g e n t f (x) = x + 15, at x = 1

(3 x^{2}) d x + (2 y - 4 x^{3} y^{- 1}) d y = 0

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