What is the derivative of f(x)=xe^{3x^2+1} ?

The derivative of f(x)=xe^{3x^2+1} is e^{3x^2+1}+6e^{3x^2+1}x^2

What is the first derivative of f(x)=xe^{3x^2+1} ?

The first derivative of f(x)=xe^{3x^2+1} is e^{3x^2+1}+6e^{3x^2+1}x^2

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derivative of

f (x) = x e^{3 x^{2} + 1}

e^{3 x^{2} + 1} + 6 e^{3 x^{2} + 1} x^{2}

Solution steps

\frac{d}{d x} (x e^{3 x^{2} + 1})

Apply the Product Rule:

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

f = x, g = e^{3 x^{2} + 1}

= \frac{d x}{d x} e^{3 x^{2} + 1} + \frac{d}{d x} (e^{3 x^{2} + 1}) x

\frac{d x}{d x} = 1

\frac{d}{d x} (e^{3 x^{2} + 1}) = e^{3 x^{2} + 1} \cdot 6 x

= 1 \cdot e^{3 x^{2} + 1} + e^{3 x^{2} + 1} \cdot 6 xx

Simplify

1 \cdot e^{3 x^{2} + 1} + e^{3 x^{2} + 1} \cdot 6 xx : e^{3 x^{2} + 1} + 6 e^{3 x^{2} + 1} x^{2}

= e^{3 x^{2} + 1} + 6 e^{3 x^{2} + 1} x^{2}

\int \frac{1}{( y + 1 ) ( y - 2 )} d y

t an g e n t f (x) = x^{2} + sin (x)

\int x^{2} - 3 x d x

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

\frac{d}{d x} (lo g_{e} (x + 1))

What is the derivative of f(x)=xe^{3x^2+1} ?
The derivative of f(x)=xe^{3x^2+1} is e^{3x^2+1}+6e^{3x^2+1}x^2
What is the first derivative of f(x)=xe^{3x^2+1} ?
The first derivative of f(x)=xe^{3x^2+1} is e^{3x^2+1}+6e^{3x^2+1}x^2