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

The derivative of sin(1/x x^3) is -xcos(1/x)+3x^2sin(1/x)

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

The first derivative of sin(1/x x^3) is -xcos(1/x)+3x^2sin(1/x)

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

g (x) = ln (x^{2}), (e, 2)

\int_{0}^{8} t + 2 t^{3} d t

(8.2) (- 10.4)

\int \frac{x + 2}{x ^{2} - 4 x} d x

\frac{\partial}{\partial y} (x^{3} + y^{2} - y x^{2} - x y)

What is the derivative of sin(1/x x^3) ?
The derivative of sin(1/x x^3) is -xcos(1/x)+3x^2sin(1/x)
What is the first derivative of sin(1/x x^3) ?
The first derivative of sin(1/x x^3) is -xcos(1/x)+3x^2sin(1/x)