What is the derivative of ye^{x^2} ?

The derivative of ye^{x^2} is (dy)/(dx)e^{x^2}+e^{x^2}*2xy

What is the first derivative of ye^{x^2} ?

The first derivative of ye^{x^2} is (dy)/(dx)e^{x^2}+e^{x^2}*2xy

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

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

Solution steps

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

Apply the Product Rule:

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

f = y, g = e^{x^{2}}

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

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

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

6 cos (3 x)

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

\int \frac{( x ^{2} - 16 )}{x ^{3}} d x

y^{^{'}} = 6 y + 2 t - 4, y (0) = 3

\int_{0}^{1} 1 - x d x

What is the derivative of ye^{x^2} ?
The derivative of ye^{x^2} is (dy)/(dx)e^{x^2}+e^{x^2}*2xy
What is the first derivative of ye^{x^2} ?
The first derivative of ye^{x^2} is (dy)/(dx)e^{x^2}+e^{x^2}*2xy