What is taylor 1/(2+5x) ?

The answer to taylor 1/(2+5x) is 1/2-5/4 x+25/8 x^2-125/16 x^3+625/32 x^4+\ldots

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taylor 1/(2+5x)

taylor

\frac{1}{2 + 5 x}

\frac{1}{2} - \frac{5}{4} x + \frac{25}{8} x^{2} - \frac{125}{16} x^{3} + \frac{625}{32} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= \frac{1}{2} + \frac{\frac{d}{d x} ( \frac{1}{2 + 5 x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{2 + 5 x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{2 + 5 x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= \frac{1}{2} + \frac{- \frac{5}{4}}{1 !} x + \frac{\frac{25}{4}}{2 !} x^{2} + \frac{- \frac{375}{8}}{3 !} x^{3} + \frac{\frac{1875}{4}}{4 !} x^{4} + \dots

Refine

= \frac{1}{2} - \frac{5}{4} x + \frac{25}{8} x^{2} - \frac{125}{16} x^{3} + \frac{625}{32} x^{4} + \dots

\frac{d y}{d x} = y^{2} - 9

\frac{d}{d x} (5 x^{3} + x^{2} + \frac{1}{x})

\frac{d}{d x} (2 (5 - 9 x)^{5})

\int \frac{1 - x}{1 - x ^{2}} d x

\int_{0}^{2} (x^{2} - 2 x) d x

What is taylor 1/(2+5x) ?
The answer to taylor 1/(2+5x) is 1/2-5/4 x+25/8 x^2-125/16 x^3+625/32 x^4+\ldots