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

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

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

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

4 y x - sin (x^{4}) \cdot 4 x^{3}

Solution steps

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

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

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

= 4 y x - sin (x^{4}) \cdot 4 x^{3}

\int \frac{1}{10 - x} d x

\frac{\partial}{\partial z} (x tan (y) - z e^{z})

\frac{d y}{d t} + 3 y = t + e^{- 2 t}

\int \frac{4}{3 2 x + 1} d x

\int (1 + sin (2 x))^{2} d x

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