What is the derivative of (8x^2+9x)e^{-7x} ?

The derivative of (8x^2+9x)e^{-7x} is -56e^{-7x}x^2-47e^{-7x}x+9e^{-7x}

What is the first derivative of (8x^2+9x)e^{-7x} ?

The first derivative of (8x^2+9x)e^{-7x} is -56e^{-7x}x^2-47e^{-7x}x+9e^{-7x}

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derivative of

(8 x^{2} + 9 x) e^{- 7 x}

- 56 e^{- 7 x} x^{2} - 47 e^{- 7 x} x + 9 e^{- 7 x}

Solution steps

\frac{d}{d x} ((8 x^{2} + 9 x) e^{- 7 x})

Apply the Product Rule:

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

f = 8 x^{2} + 9 x, g = e^{- 7 x}

= \frac{d}{d x} (8 x^{2} + 9 x) e^{- 7 x} + \frac{d}{d x} (e^{- 7 x}) (8 x^{2} + 9 x)

\frac{d}{d x} (8 x^{2} + 9 x) = 16 x + 9

\frac{d}{d x} (e^{- 7 x}) = - 7 e^{- 7 x}

= (16 x + 9) e^{- 7 x} + (- 7 e^{- 7 x}) (8 x^{2} + 9 x)

Simplify

(16 x + 9) e^{- 7 x} + (- 7 e^{- 7 x}) (8 x^{2} + 9 x) : - 56 e^{- 7 x} x^{2} - 47 e^{- 7 x} x + 9 e^{- 7 x}

= - 56 e^{- 7 x} x^{2} - 47 e^{- 7 x} x + 9 e^{- 7 x}

\frac{d y}{d x} = x^{3} (1 - y), y (0) = 3

\int_{- \infty}^{- 8} x^{- \frac{4}{3}} d x

\frac{d x}{d t} = 6 - \frac{6 x}{15}

x \to \infty lim (2 - \frac{1}{x})

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

What is the derivative of (8x^2+9x)e^{-7x} ?
The derivative of (8x^2+9x)e^{-7x} is -56e^{-7x}x^2-47e^{-7x}x+9e^{-7x}
What is the first derivative of (8x^2+9x)e^{-7x} ?
The first derivative of (8x^2+9x)e^{-7x} is -56e^{-7x}x^2-47e^{-7x}x+9e^{-7x}