What is the derivative of \sqrt[4]{sin(x^3}) ?

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

What is the first derivative of \sqrt[4]{sin(x^3}) ?

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

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int e^{c o s (6 t)} sin (6 t) d t

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = - 4, y^{^{'}} (0) = 4

\int \frac{1}{100 + 2 t} d t

x \to + 0 lim (3 x)

\frac{\partial}{\partial y} (1 + y e^{x y})

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