What is the derivative of y=2^{2x} ?

The derivative of y=2^{2x} is ln(2)*2^{2x+1}

What is the first derivative of y=2^{2x} ?

The first derivative of y=2^{2x} is ln(2)*2^{2x+1}

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derivative of

y = 2^{2 x}

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

Solution steps

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

Simplify

2^{2 x} : 4^{x}

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

Apply the derivative exponent rule:

\frac{d}{d x} (a^{x}) = a^{x} ln (a)

= 4^{x} ln (4)

Simplify

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

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

y = e^{in (x^{2} + 1)}

x \to 2 lim (\frac{3 3 x ^{2} - 2 x - 8}{( x + 2 ) ^{3}})

x \to 9 lim (x - 7)

\frac{1}{( 1 + x ^{2} )}

\int \frac{x ^{2}}{9 - x ^{2}} d x

What is the derivative of y=2^{2x} ?
The derivative of y=2^{2x} is ln(2)*2^{2x+1}
What is the first derivative of y=2^{2x} ?
The first derivative of y=2^{2x} is ln(2)*2^{2x+1}