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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

x^{2} (4 - x)^{2}

- 13 \cdot 7 e^{- t}

e^{- 3} \cdot e^{x}

- e^{- a}

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