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Question #e1fc4

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Jul 24, 2015

$\frac{d y}{d x} = (2 ln 2) 4^{sec (x)} sec (x) tan (x)$ $dy/dx=(2ln2)4^sec(x)sec(x)tan(x)$

Explanation:

Lets say we want to find the derivative of $2^{x}$ $2^x$ .
It is $(ln 2) 2^{x}$ $(ln2)2^x$

In the problem we do the same thing and remember
to use the chain rule.

$\frac{d y}{d x} = (ln 4) 4^{sec (x)} \frac{d}{d x} [sec (x)]$ $dy/dx=(ln4)4^sec(x)d/dx[sec(x)]$

$\frac{d y}{d x} = ({ln 2}^{2}) 4^{sec (x)} sec (x) tan (x)$ $dy/dx=(ln2^2)4^sec(x)sec(x)tan(x)$

$\frac{d y}{d x} = (2 ln 2) 4^{sec (x)} sec (x) tan (x)$ $dy/dx=(2ln2)4^sec(x)sec(x)tan(x)$

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