What is taylor xln(7x),1 ?

The answer to taylor xln(7x),1 is ln(7)+(ln(7)+1)(x-1)+1/2 (x-1)^2-1/6 (x-1)^3+1/12 (x-1)^4+\ldots

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taylor xln(7x),1

taylor

x ln (7 x), 1

ln (7) + (ln (7) + 1) (x - 1) + \frac{1}{2} (x - 1)^{2} - \frac{1}{6} (x - 1)^{3} + \frac{1}{12} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

= ln (7) + \frac{\frac{d}{d x} ( x ln ( 7 x ) ) ( 1 )}{1 !} (x - 1) + \frac{\frac{d ^{2}}{d x ^{2}} ( x ln ( 7 x ) ) ( 1 )}{2 !} (x - 1)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ln ( 7 x ) ) ( 1 )}{3 !} (x - 1)^{3} + \dots

Evaluate Derivatives

= ln (7) + \frac{ln ( 7 ) + 1}{1 !} (x - 1) + \frac{1}{2 !} (x - 1)^{2} + \frac{- 1}{3 !} (x - 1)^{3} + \frac{2}{4 !} (x - 1)^{4} + \dots

Refine

= ln (7) + (ln (7) + 1) (x - 1) + \frac{1}{2} (x - 1)^{2} - \frac{1}{6} (x - 1)^{3} + \frac{1}{12} (x - 1)^{4} + \dots

\int \frac{4 ln ( x )}{x ^{3}} d x

f (x) = \frac{c ^{4} - x ^{4}}{c ^{4} + x ^{4}}

\frac{d}{d y} (y^{- 2})

\int_{- 1}^{1} e^{x} - x^{2} + 1 d x

\int \frac{x ^{2}}{x ^{2} + x ^{4}} d x

What is taylor xln(7x),1 ?
The answer to taylor xln(7x),1 is ln(7)+(ln(7)+1)(x-1)+1/2 (x-1)^2-1/6 (x-1)^3+1/12 (x-1)^4+\ldots