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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= y \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})

= y (\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}}

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

Simplify

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

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

x \to 0 lim (- x^{3})

f (x) = x^{2} + 6

\int \frac{1}{y - y ln ( y )} d y

4 - 3 x

f (x) = ln (sin (x) + 4)

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