Use differentials to estimate the following: a)sqrt(36.1) b) tan 44 degrees

a) You should use the formula of linear approximation near a point x=a such that:

`L(x) = f(a) + f'(a)(x-a)`

You should notice that the function that you need to use is `f(x) = sqrt x =gt f'(x) = 1/(2sqrtx)`

You need to substitute 36.1 for x in f(x) and f'(x) such that:

`L(x) = f(36.1) + f'(36.1)(x-36.1)`

`f(36.1) = sqrt(36.1)`

`f'(36.1) = sqrt(36.1)/72.2`

`L(x) = sqrt(36.1) + sqrt(36.1)/72.2 (x - 36.1)`

`L(x) = sqrt(36.1)(1 + (x - 36.1)/72.2)`

`L(x)|_(x=36.1) = 6.008*(1) `

Hence, using linear approximation (differentials) to estimate `sqrt(36.1)` yields `L(x)|_(x=36.1) = 6.008.`

b) You should use the formula of linear approximation near a point x=a such that:

`L(x) = f(a) + f'(a)(x-a)`

You should notice that the function that you need to use is `f(x) =tan x=gt f'(x) = 1/(cos^2 x)`

You need to substitute `44^o`  for x in f(x) and f'(x) such that:

`L(x) = f(44^o) + f'(44^o)(x-44^o)`

`L(x) = tan 44^o + 1/(cos^2 44^o) (x - 44^o)`

`L(x) = 0.965 + 1.937(x - 44^o)`

`L(x)|_(x=44^o) = 0.965`

Hence, evaluating `tan 44^o`  using linear approximation yields `L(x)|_(x=44^o) = 0.965` .

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