What is maclaurin 2^x ?

The answer to maclaurin 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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maclaurin 2^x

maclaurin

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 Maclaurin 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

\frac{d}{d x} ((x + ln (x^{2}))^{s i n (x) + 2 c o s (x)})

\frac{\partial}{\partial x} (6 x^{2} y + 1)

x \to 3 lim (\frac{1}{3 x ^{2} - 27})

\frac{\partial}{\partial x} (3 sin^{2} (x) cos (x))

\int y sin (x) d x

What is maclaurin 2^x ?
The answer to maclaurin 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