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)

(\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_{0}^{1} x (x - x) d x

\frac{d}{d x} (ln (x - 2))

\int_{0}^{π} (5 e^{x} + 5 sin (x)) d x

d er i v a t i v e f (x) = \frac{1}{2 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)