What is (\partial)/(\partial x)(x^2sin(1/x)) ?

The answer to (\partial)/(\partial x)(x^2sin(1/x)) is 2xsin(1/x)-cos(1/x)

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(\partial)/(\partial x)(x^2sin(1/x))

\frac{\partial}{\partial x} (x^{2} sin (\frac{1}{x}))

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

Solution steps

\frac{\partial}{\partial x} (x^{2} sin (\frac{1}{x}))

Apply the Product Rule:

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

f = x^{2}, g = sin (\frac{1}{x})

= \frac{\partial}{\partial x} (x^{2}) sin (\frac{1}{x}) + \frac{\partial}{\partial x} (sin (\frac{1}{x})) x^{2}

\frac{\partial}{\partial x} (x^{2}) = 2 x

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

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

Simplify

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

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

\frac{1}{x ( x + 1 )}

\int \frac{3 x ^{2} + 4 x + 1}{3 x ^{3} + 6 x ^{2} + 3 x + 5} d x

y = x^{2} sin^{6} (x) + x cos^{- 4} (x)

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

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

What is (\partial)/(\partial x)(x^2sin(1/x)) ?
The answer to (\partial)/(\partial x)(x^2sin(1/x)) is 2xsin(1/x)-cos(1/x)