What is the derivative of y=4^x(2x^4) ?

The derivative of y=4^x(2x^4) is ln(2)*2^{2x+2}x^4+2^{2x+3}x^3

What is the first derivative of y=4^x(2x^4) ?

The first derivative of y=4^x(2x^4) is ln(2)*2^{2x+2}x^4+2^{2x+3}x^3

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derivative of

y = 4^{x} (2 x^{4})

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

Solution steps

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

Simplify

4^{x} \cdot 2 x^{4} : 2^{2 x + 1} x^{4}

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

Apply the Product Rule:

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

f = 2^{2 x + 1}, g = x^{4}

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

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

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

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

4 x^{3} \cdot 2^{2 x + 1} = 2^{2 x + 3} x^{3}

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

3 (e^{x} + e^{x} x)

\frac{d}{d θ} (6 cos (θ) + 8 sin (θ))

\frac{d y}{d x} = \frac{( 2 x sec ( \frac{y}{x} ) + y )}{x}

\frac{d y}{d x} + \frac{8 y}{x + 2} = (x + 2)^{6}

What is the derivative of y=4^x(2x^4) ?
The derivative of y=4^x(2x^4) is ln(2)*2^{2x+2}x^4+2^{2x+3}x^3
What is the first derivative of y=4^x(2x^4) ?
The first derivative of y=4^x(2x^4) is ln(2)*2^{2x+2}x^4+2^{2x+3}x^3