What is maclaurin a^x ?

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

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maclaurin a^x

maclaurin

a^{x}

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

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

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

\frac{d}{d x} (- cos (4 x))

\frac{\partial}{\partial x} (ln (x^{2} + x))

\frac{\partial}{\partial x} (4 x - 4 - y^{2})

\frac{85 t}{99 t - 85}

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

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