What is taylor x^{1/2},4 ?

The answer to taylor x^{1/2},4 is 2+1/4 (x-4)-1/64 (x-4)^2+1/512 (x-4)^3-5/16384 (x-4)^4+\ldots

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taylor x^{1/2},4

taylor

x^{\frac{1}{2}}, 4

2 + \frac{1}{4} (x - 4) - \frac{1}{64} (x - 4)^{2} + \frac{1}{512} (x - 4)^{3} - \frac{5}{16384} (x - 4)^{4} + \dots

Solution steps

Apply the Taylor Formula

= 2 + \frac{\frac{d}{d x} ( x ^{\frac{1}{2}} ) ( 4 )}{1 !} (x - 4) + \frac{\frac{d ^{2}}{d x ^{2}} ( x ^{\frac{1}{2}} ) ( 4 )}{2 !} (x - 4)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ^{\frac{1}{2}} ) ( 4 )}{3 !} (x - 4)^{3} + \dots

Evaluate Derivatives

= 2 + \frac{\frac{1}{4}}{1 !} (x - 4) + \frac{- \frac{1}{32}}{2 !} (x - 4)^{2} + \frac{\frac{3}{256}}{3 !} (x - 4)^{3} + \frac{- \frac{15}{2048}}{4 !} (x - 4)^{4} + \dots

Refine

= 2 + \frac{1}{4} (x - 4) - \frac{1}{64} (x - 4)^{2} + \frac{1}{512} (x - 4)^{3} - \frac{5}{16384} (x - 4)^{4} + \dots

\frac{d}{d x} (2 x + \frac{π}{2})

x \to 0 + lim ((1 - x) sin (\frac{1}{x}))

\int 36 sec^{4} (x) d x

\int_{- 9}^{9} 1 - \frac{x ^{2}}{81} d x

f (x) = 2 x + 8 x

What is taylor x^{1/2},4 ?
The answer to taylor x^{1/2},4 is 2+1/4 (x-4)-1/64 (x-4)^2+1/512 (x-4)^3-5/16384 (x-4)^4+\ldots