What is the derivative of-cos(\sqrt[3]{x}-2+pi/2) ?

The derivative of-cos(\sqrt[3]{x}-2+pi/2) is (sin(\sqrt[3]{x}-2+pi/2))/(3x^{2/3)}

What is the first derivative of-cos(\sqrt[3]{x}-2+pi/2) ?

The first derivative of-cos(\sqrt[3]{x}-2+pi/2) is (sin(\sqrt[3]{x}-2+pi/2))/(3x^{2/3)}

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\frac{d}{d x} (- cos (3 x - 2 + \frac{π}{2}))

\frac{sin ( 3 x - 2 + \frac{π}{2} )}{3 x ^{\frac{2}{3}}}

Solution steps

\frac{d}{d x} (- cos (3 x - 2 + \frac{π}{2}))

Take the constant out:

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

= - \frac{d}{d x} (cos (3 x - 2 + \frac{π}{2}))

Apply the chain rule:

- sin (3 x - 2 + \frac{π}{2}) \frac{d}{d x} (3 x - 2 + \frac{π}{2})

= - sin (3 x - 2 + \frac{π}{2}) \frac{d}{d x} (3 x - 2 + \frac{π}{2})

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

= - (- sin (3 x - 2 + \frac{π}{2}) \frac{1}{3 x ^{\frac{2}{3}}})

Simplify

- (- sin (3 x - 2 + \frac{π}{2}) \frac{1}{3 x ^{\frac{2}{3}}}) : \frac{sin ( 3 x - 2 + \frac{π}{2} )}{3 x ^{\frac{2}{3}}}

= \frac{sin ( 3 x - 2 + \frac{π}{2} )}{3 x ^{\frac{2}{3}}}

- (\frac{a}{2})^{2} y^{^{''}} - r y^{^{'}} + ry = 0

\int - 28 x d x

\int x^{3} + sinh (x) d x

y^{^{''}} + 5 y^{^{'}} + 6 y = t e^{- t}

x \to 3 lim (9 - x^{2})

What is the derivative of-cos(\sqrt[3]{x}-2+pi/2) ?
The derivative of-cos(\sqrt[3]{x}-2+pi/2) is (sin(\sqrt[3]{x}-2+pi/2))/(3x^{2/3)}
What is the first derivative of-cos(\sqrt[3]{x}-2+pi/2) ?
The first derivative of-cos(\sqrt[3]{x}-2+pi/2) is (sin(\sqrt[3]{x}-2+pi/2))/(3x^{2/3)}