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Chain Rule of Differentiation

y = 2 x + 1...

Question

$y = (\frac{2^{x + 1}}{1 + 4^{x}})$ Find $\frac{d y}{d x}$

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Solution

$y = \frac{2^{x + 1}}{1 + 4^{x}} \Rightarrow \frac{d y}{d x} = \frac{(1 + 4^{x} .2 {.2}^{x} l o g 2 - 2^{x + 1} 4^{x} l o g 4)}{(1 + 4^{x})^{2}} = \frac{(1 + 4^{x}) 2^{(} x + 1) l o g 2 - 2^{3 x + 2} l o g 2}{(1 + 4^{x})^{2}}$

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