What is the derivative of 4^xln(4) ?

The derivative of 4^xln(4) is ln^2(2)*2^{2x+2}

What is the first derivative of 4^xln(4) ?

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

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\frac{d}{d x} (4^{x} ln (4))

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

Solution steps

\frac{d}{d x} (4^{x} ln (4))

Simplify

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

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

Take the constant out:

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

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

2^{2 x + 1} = e^{(2 x + 1) l n (2)}

= ln (2) \frac{d}{d x} (e^{(2 x + 1) l n (2)})

Apply the chain rule:

e^{(2 x + 1) l n (2)} \frac{d}{d x} ((2 x + 1) ln (2))

= e^{(2 x + 1) l n (2)} \frac{d}{d x} ((2 x + 1) ln (2))

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

= ln (2) e^{(2 x + 1) l n (2)} \cdot 2 ln (2)

Simplify

ln (2) e^{(2 x + 1) l n (2)} \cdot 2 ln (2) : ln^{2} (2) \cdot 2^{2 x + 2}

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

y^{^{'}} (y - x^{2}) = - 2 y

\frac{\partial}{\partial x} (ln (x^{4} + y^{2}))

y^{^{'}} = 7 x + y, y (0) = 8

y = x^{3}, y = 3 x + 2

\frac{\partial}{\partial x} (2 x y + y^{3})

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