What is maclaurin f(x)=x^4sin(4x) ?

The answer to maclaurin f(x)=x^4sin(4x) is 4x^5-32/3 x^7+128/15 x^9+\ldots

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maclaurin f(x)=x^4sin(4x)

maclaurin

f (x) = x^{4} sin (4 x)

4 x^{5} - \frac{32}{3} x^{7} + \frac{128}{15} x^{9} + \dots

Solution steps

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d x} ( x ^{4} sin ( 4 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( x ^{4} sin ( 4 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ^{4} sin ( 4 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{0}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{480}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 53760}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{3096576}{9 !} x^{9} + \frac{0}{10 !} x^{10} + \dots

Refine

= 4 x^{5} - \frac{32}{3} x^{7} + \frac{128}{15} x^{9} + \dots

x \to 2 lim (x (2))

\int_{3}^{14} x d x

\int cos^{2} (u) sin (u) d u

y^{^{''}} + 9 y = cos (3 x) + sin (3 x)

y^{^{'}} = \frac{6 x ^{4} e ^{\frac{y}{x}} + x ^{2} y ^{2}}{x ^{3} y}

What is maclaurin f(x)=x^4sin(4x) ?
The answer to maclaurin f(x)=x^4sin(4x) is 4x^5-32/3 x^7+128/15 x^9+\ldots