What is the derivative of 1/x sin^3(6x) ?

The derivative of 1/x sin^3(6x) is -(sin^3(6x)}{x^2}+\frac{18sin^2(6x)cos(6x))/x

What is the first derivative of 1/x sin^3(6x) ?

The first derivative of 1/x sin^3(6x) is -(sin^3(6x)}{x^2}+\frac{18sin^2(6x)cos(6x))/x

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derivative of

\frac{1}{x} sin^{3} (6 x)

- \frac{sin ^{3} ( 6 x )}{x ^{2}} + \frac{18 sin ^{2} ( 6 x ) cos ( 6 x )}{x}

Solution steps

\frac{d}{d x} (\frac{1}{x} sin^{3} (6 x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = \frac{1}{x}, g = sin^{3} (6 x)

= \frac{d}{d x} (\frac{1}{x}) sin^{3} (6 x) + \frac{d}{d x} (sin^{3} (6 x)) \frac{1}{x}

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

\frac{d}{d x} (sin^{3} (6 x)) = 18 sin^{2} (6 x) cos (6 x)

= (- \frac{1}{x ^{2}}) sin^{3} (6 x) + 18 sin^{2} (6 x) cos (6 x) \frac{1}{x}

Simplify

(- \frac{1}{x ^{2}}) sin^{3} (6 x) + 18 sin^{2} (6 x) cos (6 x) \frac{1}{x} : - \frac{sin ^{3} ( 6 x )}{x ^{2}} + \frac{18 sin ^{2} ( 6 x ) cos ( 6 x )}{x}

= - \frac{sin ^{3} ( 6 x )}{x ^{2}} + \frac{18 sin ^{2} ( 6 x ) cos ( 6 x )}{x}

7 e^{- 4 x}

\int - 0.6 x^{2} + 80 d x

\int θ sec^{2} (θ) d θ

1 + 2 x

\int \frac{1}{t ^{2} - 1} d t

What is the derivative of 1/x sin^3(6x) ?
The derivative of 1/x sin^3(6x) is -(sin^3(6x)}{x^2}+\frac{18sin^2(6x)cos(6x))/x
What is the first derivative of 1/x sin^3(6x) ?
The first derivative of 1/x sin^3(6x) is -(sin^3(6x)}{x^2}+\frac{18sin^2(6x)cos(6x))/x