What is the derivative of y=\sqrt[4]{sin(3x-2)} ?

The derivative of y=\sqrt[4]{sin(3x-2)} is (3cos(3x-2))/(4sin^{3/4)(3x-2)}

What is the first derivative of y=\sqrt[4]{sin(3x-2)} ?

The first derivative of y=\sqrt[4]{sin(3x-2)} is (3cos(3x-2))/(4sin^{3/4)(3x-2)}

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derivative of

\frac{3 cos ( 3 x - 2 )}{4 sin ^{\frac{3}{4}} ( 3 x - 2 )}

Solution steps

Apply the chain rule:

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

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

\frac{d}{d x} (sin (3 x - 2)) = cos (3 x - 2) \cdot 3

= \frac{1}{4 ( sin ( 3 x - 2 ) ) ^{\frac{3}{4}}} cos (3 x - 2) \cdot 3

Simplify

\frac{1}{4 sin ^{\frac{3}{4}} ( 3 x - 2 )} cos (3 x - 2) \cdot 3 : \frac{3 cos ( 3 x - 2 )}{4 sin ^{\frac{3}{4}} ( 3 x - 2 )}

= \frac{3 cos ( 3 x - 2 )}{4 sin ^{\frac{3}{4}} ( 3 x - 2 )}

\int csc^{2} (6 x + 3) d x

\frac{d}{d x} ((\frac{x}{2})^{4})

\frac{d}{d x} (a ln (2 x + b))

f (x) = (ln (x))^{3}

What is the derivative of y=\sqrt[4]{sin(3x-2)} ?
The derivative of y=\sqrt[4]{sin(3x-2)} is (3cos(3x-2))/(4sin^{3/4)(3x-2)}
What is the first derivative of y=\sqrt[4]{sin(3x-2)} ?
The first derivative of y=\sqrt[4]{sin(3x-2)} is (3cos(3x-2))/(4sin^{3/4)(3x-2)}