What is the derivative of 3x^24^{-x} ?

The derivative of 3x^24^{-x} is 3(2^{1-2x}x-ln(2)*2^{-2x+1}x^2)

What is the first derivative of 3x^24^{-x} ?

The first derivative of 3x^24^{-x} is 3(2^{1-2x}x-ln(2)*2^{-2x+1}x^2)

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\frac{d}{d x} (3 x^{2} 4^{- x})

3 (2^{1 - 2 x} x - ln (2) \cdot 2^{- 2 x + 1} x^{2})

Solution steps

\frac{d}{d x} (3 x^{2} \cdot 4^{- x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d x} (x^{2} \cdot 4^{- x})

Apply the Product Rule:

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

f = x^{2}, g = 4^{- x}

= 3 (\frac{d}{d x} (x^{2}) 4^{- x} + \frac{d}{d x} (4^{- x}) x^{2})

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

\frac{d}{d x} (4^{- x}) = - ln (2) \cdot 2^{- 2 x + 1}

= 3 (2 x \cdot 4^{- x} + (- ln (2) \cdot 2^{- 2 x + 1}) x^{2})

Simplify

3 (2 x \cdot 4^{- x} + (- ln (2) \cdot 2^{- 2 x + 1}) x^{2}) : 3 (2^{1 - 2 x} x - ln (2) \cdot 2^{- 2 x + 1} x^{2})

= 3 (2^{1 - 2 x} x - ln (2) \cdot 2^{- 2 x + 1} x^{2})

t \frac{d y}{d t} = t^{3} + 8 t^{3} y

\int 9 cos (θ) d θ

\frac{d}{d x} (- 2 x^{2} + 7 x + 5)

n = 0 \sum \infty ln (1 + n)

\frac{\partial}{\partial y} (x^{y} - y^{x})

What is the derivative of 3x^24^{-x} ?
The derivative of 3x^24^{-x} is 3(2^{1-2x}x-ln(2)*2^{-2x+1}x^2)
What is the first derivative of 3x^24^{-x} ?
The first derivative of 3x^24^{-x} is 3(2^{1-2x}x-ln(2)*2^{-2x+1}x^2)