The answer to taylor 2^x is 1+ln(2)x+(ln^2(2))/2 x^2+(ln^3(2))/6 x^3+(ln^4(2))/(24)x^4+\ldots

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taylor 2^x

Solution

taylor

2^{x}

1 + ln (2) x + \frac{ln ^{2} ( 2 )}{2} x^{2} + \frac{ln ^{3} ( 2 )}{6} x^{3} + \frac{ln ^{4} ( 2 )}{24} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

= 1 + ln (2) x + \frac{ln ^{2} ( 2 )}{2} x^{2} + \frac{ln ^{3} ( 2 )}{6} x^{3} + \frac{ln ^{4} ( 2 )}{24} x^{4} + \dots

\int \frac{x ^{2} + 48 x - 16}{x ^{3} - 16 x} d x

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

\frac{d y}{d t} = - \frac{y}{t ^{2}}

f (x) = 6 x - x^{2}, at x = 4

\frac{d}{d x} ((x - 5) sin (x))

What is taylor 2^x ?
The answer to taylor 2^x is 1+ln(2)x+(ln^2(2))/2 x^2+(ln^3(2))/6 x^3+(ln^4(2))/(24)x^4+\ldots