What is the derivative of (sqrt(2)^x) ?

The derivative of (sqrt(2)^x) is 1/2 ln(2)e^{(ln(2)x)/2}

What is the first derivative of (sqrt(2)^x) ?

The first derivative of (sqrt(2)^x) is 1/2 ln(2)e^{(ln(2)x)/2}

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

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

Solution steps

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

Simplify

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

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

Apply exponent rule:

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

2^{\frac{x}{2}} = e^{\frac{x}{2} l n (2)}

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

Apply the chain rule:

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

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

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

4 x - x^{2}

\int cos (9 x) sin (5 x) d x

y^{3} - (18 x + 6) + 3 x y^{2} \frac{d y}{d x} = 0

\frac{d}{d x} ((\frac{1 + sin ( 3 x )}{3 - 2 x})^{- 1})

n = 0 \sum \infty (1 + x)^{n}

What is the derivative of (sqrt(2)^x) ?
The derivative of (sqrt(2)^x) is 1/2 ln(2)e^{(ln(2)x)/2}
What is the first derivative of (sqrt(2)^x) ?
The first derivative of (sqrt(2)^x) is 1/2 ln(2)e^{(ln(2)x)/2}