What is the derivative of f(x)=3x^2e^x+e^x(x^3+9) ?

The derivative of f(x)=3x^2e^x+e^x(x^3+9) is e^xx^3+6e^xx^2+6e^xx+9e^x

What is the first derivative of f(x)=3x^2e^x+e^x(x^3+9) ?

The first derivative of f(x)=3x^2e^x+e^x(x^3+9) is e^xx^3+6e^xx^2+6e^xx+9e^x

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derivative of

f (x) = 3 x^{2} e^{x} + e^{x} (x^{3} + 9)

e^{x} x^{3} + 6 e^{x} x^{2} + 6 e^{x} x + 9 e^{x}

Solution steps

\frac{d}{d x} (3 x^{2} e^{x} + e^{x} (x^{3} + 9))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (3 x^{2} e^{x}) + \frac{d}{d x} (e^{x} (x^{3} + 9))

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

\frac{d}{d x} (e^{x} (x^{3} + 9)) = e^{x} (x^{3} + 9) + 3 x^{2} e^{x}

= 3 (2 x e^{x} + e^{x} x^{2}) + e^{x} (x^{3} + 9) + 3 x^{2} e^{x}

Simplify

3 (2 x e^{x} + e^{x} x^{2}) + e^{x} (x^{3} + 9) + 3 x^{2} e^{x} : e^{x} x^{3} + 6 e^{x} x^{2} + 6 e^{x} x + 9 e^{x}

= e^{x} x^{3} + 6 e^{x} x^{2} + 6 e^{x} x + 9 e^{x}

y \frac{d y}{d x} = \frac{4}{5} x

f (x) = 4 sec (x), at x = \frac{7 π}{6}

\int \frac{2 x ^{2}}{( x - 1 ) ^{2} + 4} d x

y = (x + 2) (x - 1) (x - 3)

\int t sec^{2} (9 t) d t

What is the derivative of f(x)=3x^2e^x+e^x(x^3+9) ?
The derivative of f(x)=3x^2e^x+e^x(x^3+9) is e^xx^3+6e^xx^2+6e^xx+9e^x
What is the first derivative of f(x)=3x^2e^x+e^x(x^3+9) ?
The first derivative of f(x)=3x^2e^x+e^x(x^3+9) is e^xx^3+6e^xx^2+6e^xx+9e^x