What is the derivative of 0.6x^34^x ?

The derivative of 0.6x^34^x is 0.6(3*4^xx^2+0.69314…*2^{2x+1}x^3)

What is the first derivative of 0.6x^34^x ?

The first derivative of 0.6x^34^x is 0.6(3*4^xx^2+0.69314…*2^{2x+1}x^3)

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\frac{d}{d x} (0.6 x^{3} 4^{x})

0.6 (3 \cdot 4^{x} x^{2} + 0.69314 \dots \cdot 2^{2 x + 1} x^{3})

Solution steps

\frac{d}{d x} (0.6 x^{3} \cdot 4^{x})

Take the constant out:

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

= 0.6 \frac{d}{d x} (x^{3} \cdot 4^{x})

Apply the Product Rule:

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

f = x^{3}, g = 4^{x}

= 0.6 (\frac{d}{d x} (x^{3}) 4^{x} + \frac{d}{d x} (4^{x}) x^{3})

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

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

= 0.6 (3 x^{2} \cdot 4^{x} + ln (2) \cdot 2^{2 x + 1} x^{3})

Refine to a decimal form

= 0.6 (3 \cdot 4^{x} x^{2} + 0.69314 \dots \cdot 2^{2 x + 1} x^{3})

y = 3 x^{2}, y = 2, x = 0, x \geq 0

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

\frac{x}{2 + x ^{2}}

(3 x + 7) x^{2}

\frac{d}{d x} (- \frac{32}{x ^{3}} + \frac{4}{x ^{2}})

What is the derivative of 0.6x^34^x ?
The derivative of 0.6x^34^x is 0.6(3*4^xx^2+0.69314…*2^{2x+1}x^3)
What is the first derivative of 0.6x^34^x ?
The first derivative of 0.6x^34^x is 0.6(3*4^xx^2+0.69314…*2^{2x+1}x^3)