What is (\partial)/(\partial x)(2x^4y+cos(x^3)) ?

The answer to (\partial)/(\partial x)(2x^4y+cos(x^3)) is 8yx^3-sin(x^3)*3x^2

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(\partial)/(\partial x)(2x^4y+cos(x^3))

\frac{\partial}{\partial x} (2 x^{4} y + cos (x^{3}))

8 y x^{3} - sin (x^{3}) \cdot 3 x^{2}

Solution steps

\frac{\partial}{\partial x} (2 x^{4} y + cos (x^{3}))

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (2 x^{4} y) + \frac{\partial}{\partial x} (cos (x^{3}))

\frac{\partial}{\partial x} (2 x^{4} y) = 8 y x^{3}

\frac{\partial}{\partial x} (cos (x^{3})) = - sin (x^{3}) \cdot 3 x^{2}

= 8 y x^{3} - sin (x^{3}) \cdot 3 x^{2}

\int_{7}^{8} \frac{1}{x ^{2} - 1} d x

\int \frac{9}{x ( x - 9 ) ^{2}} d x

\int (tan^{2} (x) 4) d x

\frac{d}{d x} (e^{2 x^{6}} - 1)

\frac{\partial}{\partial y} (ln (x^{6} + y^{7}))

What is (\partial)/(\partial x)(2x^4y+cos(x^3)) ?
The answer to (\partial)/(\partial x)(2x^4y+cos(x^3)) is 8yx^3-sin(x^3)*3x^2