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

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

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

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

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

Solution steps

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

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

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

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

x \to 1 lim (x^{\frac{8}{1 - x}})

\frac{\partial}{\partial x} ((x^{3} + y^{3})^{\frac{1}{3}})

\int \frac{1}{x 9 - 4 x ^{2}} d x

\int 2 y - 2 d y

\int_{0}^{\infty} 5 e^{- 2 x} d x

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