What is the derivative of 1/(\sqrt[3]{(9x+1)^2)} ?

The derivative of 1/(\sqrt[3]{(9x+1)^2)} is -(6(9x+1))/(((9x+1)^2)^{4/3)}

What is the first derivative of 1/(\sqrt[3]{(9x+1)^2)} ?

The first derivative of 1/(\sqrt[3]{(9x+1)^2)} is -(6(9x+1))/(((9x+1)^2)^{4/3)}

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derivative of

- \frac{6 ( 9 x + 1 )}{( ( 9 x + 1 ) ^{2} ) ^{\frac{4}{3}}}

Solution steps

Apply exponent rule:

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

= \frac{d}{d x} (((9 x + 1)^{2})^{- \frac{1}{3}})

Apply the chain rule:

- \frac{1}{3 ( ( 9 x + 1 ) ^{2} ) ^{\frac{4}{3}}} \frac{d}{d x} ((9 x + 1)^{2})

= - \frac{1}{3 ( ( 9 x + 1 ) ^{2} ) ^{\frac{4}{3}}} \frac{d}{d x} ((9 x + 1)^{2})

\frac{d}{d x} ((9 x + 1)^{2}) = 18 (9 x + 1)

= - \frac{1}{3 ( ( 9 x + 1 ) ^{2} ) ^{\frac{4}{3}}} \cdot 18 (9 x + 1)

Simplify

- \frac{1}{3 ( ( 9 x + 1 ) ^{2} ) ^{\frac{4}{3}}} \cdot 18 (9 x + 1) : - \frac{6 ( 9 x + 1 )}{( ( 9 x + 1 ) ^{2} ) ^{\frac{4}{3}}}

= - \frac{6 ( 9 x + 1 )}{( ( 9 x + 1 ) ^{2} ) ^{\frac{4}{3}}}

x \to \infty lim (x - 4)

\frac{d}{d x} (- e^{c o s (x)} sin (x))

y = 3^{1 - x}

x, \frac{1}{x}, x = 3

\int sin (2 x + \frac{π}{2}) d x

What is the derivative of 1/(\sqrt[3]{(9x+1)^2)} ?
The derivative of 1/(\sqrt[3]{(9x+1)^2)} is -(6(9x+1))/(((9x+1)^2)^{4/3)}
What is the first derivative of 1/(\sqrt[3]{(9x+1)^2)} ?
The first derivative of 1/(\sqrt[3]{(9x+1)^2)} is -(6(9x+1))/(((9x+1)^2)^{4/3)}