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

The answer to (\partial)/(\partial x)(2x^3y+sin(x^4)) is 6yx^2+cos(x^4)*4x^3

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

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

6 y x^{2} + cos (x^{4}) \cdot 4 x^{3}

Solution steps

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

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^{4}))

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

\frac{\partial}{\partial x} (sin (x^{4})) = cos (x^{4}) \cdot 4 x^{3}

= 6 y x^{2} + cos (x^{4}) \cdot 4 x^{3}

\int \frac{- 4 x}{e ^{x^{2}}} d x

\frac{d}{d y} (3 x e^{- \frac{1}{y ^{3}}} - 20 x^{3})

f (x) = x - \frac{1}{x} + ln (x), at x = 1

y = 16 x sin (x), (\frac{π}{2}, 8 π)

\frac{\partial}{\partial y} (x ln (y + z))

What is (\partial)/(\partial x)(2x^3y+sin(x^4)) ?
The answer to (\partial)/(\partial x)(2x^3y+sin(x^4)) is 6yx^2+cos(x^4)*4x^3