What is the derivative of (yx^2) ?

The derivative of (yx^2) is (dy)/(dx)x^2+2xy

What is the first derivative of (yx^2) ?

The first derivative of (yx^2) is (dy)/(dx)x^2+2xy

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\frac{d}{d x} ((y) x^{2})

\frac{d y}{d x} x^{2} + 2 x y

Solution steps

\frac{d}{d x} ((y) x^{2})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y, g = x^{2}

= \frac{d y}{d x} x^{2} + \frac{d}{d x} (x^{2}) y

\frac{d}{d x} (x^{2}) = 2 x

= \frac{d y}{d x} x^{2} + 2 x y

\frac{\partial}{\partial x} (x z t^{2} y + 1)

f (x) = e^{3 e^{x}}

(5.4) 8 x^{2} - 9 y^{3} = - 376

f (x) = 7 x^{4} + 13 x^{3} - 22 x^{2} + 5 x - 3

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

What is the derivative of (yx^2) ?
The derivative of (yx^2) is (dy)/(dx)x^2+2xy
What is the first derivative of (yx^2) ?
The first derivative of (yx^2) is (dy)/(dx)x^2+2xy