What is (\partial)/(\partial x)(2ysin(x)-x^3+ln(y)) ?

The answer to (\partial)/(\partial x)(2ysin(x)-x^3+ln(y)) is -3x^2+2ycos(x)

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(\partial)/(\partial x)(2ysin(x)-x^3+ln(y))

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

- 3 x^{2} + 2 y cos (x)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

\frac{\partial}{\partial x} (ln (y)) = 0

= 2 y cos (x) - 3 x^{2} + 0

Simplify

= - 3 x^{2} + 2 y cos (x)

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

\int 5 x + x d x

\frac{x ( x + 1 ) d y}{d x} + x y = 1, y (e) = 1

y = \frac{4 x - 7 - x}{5}

\int (cos (x))^{2} d x

What is (\partial)/(\partial x)(2ysin(x)-x^3+ln(y)) ?
The answer to (\partial)/(\partial x)(2ysin(x)-x^3+ln(y)) is -3x^2+2ycos(x)