The terms of the Maclaurin series for [tex]f(x)=e^x sin(5x)[/tex] through degree four are [tex]5x+5x^2-\frac{55}{3}x^3-20x^4[/tex]
The function and it's derivatives, up to the fourth are
[tex]f(x)=e^xsin(5x)\\f'(x)=e^x(sin(5x)+5cos(5x))\\f''(x)=2e^x(5cos(5x)-12sin(5x))\\f'''(x)=-2e^x(37sin(5x)+55cos(5x))\\f^{iv}(x)=4e^x(119sin(5x)-120cos(5x))[/tex]
since we are dealing with the Maclaurin series, we will evaluate each of them at [tex]x=0[/tex] to get
[tex]f(0)=0\\f'(0)=5\\f''(0)=10\\f'''(0)=-110\\f^{iv}(0)=-480[/tex]
The general form of the Maclaurin series, up to degree four, is given by
[tex]f(x)=\frac{f(0)}{0!}x^0+\frac{f'(0)}{1!}x^1+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\frac{f^{iv}(0)}{4!}x^4\\\\f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\frac{f^{iv}(0)}{4!}x^4[/tex]
Substituting the previous values into this series formula, and simplifying, we get
[tex]e^x sin(5x)=0+5x+\frac{10}{2}x^2-\frac{110}{6}x^3-\frac{480}{24}x^4\\e^x sin(5x)=5x+5x^2-\frac{55}{3}x^3-20x^4[/tex]
Learn more about the Maclaurin series here: https://brainly.com/question/7846182
