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Question

# Find the linearization L(x) of ƒ(x) at x = a. ƒ(x) = tan x, a = π

Solutions

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To find the required linearization, we need to find $f'(\pi)$, where $f(x)=\tan x$

$f'(x)=\dfrac{d\left(\tan x\right)}{dx}$

Remember that: the derivative of $\tan x$ is $\sec^2x$

$f'(x)=\sec^2x$

Substitute $x=\pi$

$f'(\pi)=\sec^2\pi$

$f'(\pi)=(-1)^2=1$

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