What is the derivative of y=Ax^2+Be^{2x} ?

The derivative of y=Ax^2+Be^{2x} is 2Ax+Be^{2x}*2

What is the first derivative of y=Ax^2+Be^{2x} ?

The first derivative of y=Ax^2+Be^{2x} is 2Ax+Be^{2x}*2

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derivative of

y = A x^{2} + B e^{2 x}

2 A x + B e^{2 x} \cdot 2

Solution steps

\frac{d}{d x} (A x^{2} + B e^{2 x})

Apply the Sum/Difference Rule:

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

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

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

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

= 2 A x + B e^{2 x} \cdot 2

y^{^{'}} = 2 x y, y (0) = 1

(x y) d x + (x^{2} + y^{2}) d y = 0

2 (x - 6)^{3}

z \to 0 lim (z^{3} cos (\frac{1}{z}))

What is the derivative of y=Ax^2+Be^{2x} ?
The derivative of y=Ax^2+Be^{2x} is 2Ax+Be^{2x}*2
What is the first derivative of y=Ax^2+Be^{2x} ?
The first derivative of y=Ax^2+Be^{2x} is 2Ax+Be^{2x}*2