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

The answer to (\partial)/(\partial x)(x^3+x^2*x^3-2y^2) is 5x^4+3x^2

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(\partial)/(\partial x)(x^3+x^2*x^3-2y^2)

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

5 x^{4} + 3 x^{2}

Solution steps

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

Simplify

x^{3} + x^{2} x^{3} - 2 y^{2} : x^{3} + x^{5} - 2 y^{2}

= \frac{\partial}{\partial x} (x^{3} + x^{5} - 2 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{3}) + \frac{\partial}{\partial x} (x^{5}) - \frac{\partial}{\partial x} (2 y^{2})

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

\frac{\partial}{\partial x} (x^{5}) = 5 x^{4}

\frac{\partial}{\partial x} (2 y^{2}) = 0

= 3 x^{2} + 5 x^{4} - 0

Simplify

= 5 x^{4} + 3 x^{2}

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

\frac{d}{d x} (- 16)

\frac{d}{d x} (5^{t a n (5 x)})

(- 5, 3), (2, - 3)

(x^{4} (x - 3)^{3})^{^{'}}

What is (\partial)/(\partial x)(x^3+x^2*x^3-2y^2) ?
The answer to (\partial)/(\partial x)(x^3+x^2*x^3-2y^2) is 5x^4+3x^2