What is (\partial)/(\partial x)(2x^3y+sin(x^2)) ?

The answer to (\partial)/(\partial x)(2x^3y+sin(x^2)) is 6yx^2+cos(x^2)*2x

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(\partial)/(\partial x)(2x^3y+sin(x^2))

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

6 y x^{2} + cos (x^{2}) \cdot 2 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (2 x^{3} y) = 6 y x^{2}

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

= 6 y x^{2} + cos (x^{2}) \cdot 2 x

f (x) = \frac{4 x}{x ^{2} - 5}, (3, 3)

x \to 7 lim (\frac{x + 2}{x - 7})

\int \frac{e ^{3 x} + 1}{e ^{2 x}} d x

\int_{0}^{1} \frac{x - 6}{x ^{2} - 5 x + 6} d x

\frac{d y}{d x} - e^{8 x + y} = 0

What is (\partial)/(\partial x)(2x^3y+sin(x^2)) ?
The answer to (\partial)/(\partial x)(2x^3y+sin(x^2)) is 6yx^2+cos(x^2)*2x