What is the derivative of 6/(2^x) ?

The derivative of 6/(2^x) is -3ln(2)*2^{1-x}

What is the first derivative of 6/(2^x) ?

The first derivative of 6/(2^x) is -3ln(2)*2^{1-x}

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\frac{d}{d x} (\frac{6}{2 ^{x}})

- 3 ln (2) \cdot 2^{1 - x}

Solution steps

\frac{d}{d x} (\frac{6}{2 ^{x}})

Simplify

\frac{6}{2 ^{x}} : 3 \cdot 2^{1 - x}

= \frac{d}{d x} (3 \cdot 2^{1 - x})

Take the constant out:

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

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

Apply exponent rule:

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

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

= 3 \frac{d}{d x} (e^{(1 - x) l n (2)})

Apply the chain rule:

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

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

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

= 3 e^{(1 - x) l n (2)} (- ln (2))

Simplify

3 e^{(1 - x) l n (2)} (- ln (2)) : - 3 ln (2) \cdot 2^{1 - x}

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

\frac{d}{d x} (\frac{x}{x + 5})

x \frac{d y}{d x} = y (y + 1)

f (x) = \frac{6 x}{x + 2}, (4, 4)

f (x) = - \frac{1}{6} (x^{- 6} + x^{12})

\frac{x ^{2} - 36}{x - 6}

What is the derivative of 6/(2^x) ?
The derivative of 6/(2^x) is -3ln(2)*2^{1-x}
What is the first derivative of 6/(2^x) ?
The first derivative of 6/(2^x) is -3ln(2)*2^{1-x}