What is the derivative of y=e^x(3-5x+3x^2) ?

The derivative of y=e^x(3-5x+3x^2) is 3e^xx^2+e^xx-2e^x

What is the first derivative of y=e^x(3-5x+3x^2) ?

The first derivative of y=e^x(3-5x+3x^2) is 3e^xx^2+e^xx-2e^x

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derivative of

y = e^{x} (3 - 5 x + 3 x^{2})

3 e^{x} x^{2} + e^{x} x - 2 e^{x}

Solution steps

\frac{d}{d x} (e^{x} (3 - 5 x + 3 x^{2}))

Apply the Product Rule:

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

f = e^{x}, g = 3 - 5 x + 3 x^{2}

= \frac{d}{d x} (e^{x}) (3 - 5 x + 3 x^{2}) + \frac{d}{d x} (3 - 5 x + 3 x^{2}) e^{x}

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

\frac{d}{d x} (3 - 5 x + 3 x^{2}) = 6 x - 5

= e^{x} (3 - 5 x + 3 x^{2}) + (6 x - 5) e^{x}

Simplify

e^{x} (3 - 5 x + 3 x^{2}) + (6 x - 5) e^{x} : 3 e^{x} x^{2} + e^{x} x - 2 e^{x}

= 3 e^{x} x^{2} + e^{x} x - 2 e^{x}

n = 0 \sum \infty n^{2} + 1

\int cos (x) 4 - sin^{2} (x) d x

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

\int x^{9} d x

(1 - y) \cdot e^{y} \cdot y^{^{'}} + \frac{y ^{2}}{x \cdot ln ( x )} = 0

What is the derivative of y=e^x(3-5x+3x^2) ?
The derivative of y=e^x(3-5x+3x^2) is 3e^xx^2+e^xx-2e^x
What is the first derivative of y=e^x(3-5x+3x^2) ?
The first derivative of y=e^x(3-5x+3x^2) is 3e^xx^2+e^xx-2e^x