The general formula for the Taylor series about x=a for a given function is

\(\displaystyle \sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}\)

We must find the zeroth, first, and second derivative of the function (for n=0, 1, and 2). The zeroth derivative is just the function itself.

\(\displaystyle f'(x)=20x+1\)

\(\displaystyle f''(x)=20\)

The derivatives were found using the following rule:
\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}\)

Now, follow the above formula to write out the first three terms:

\(\displaystyle \frac{11(x-1)^0}{0!}+\frac{21(x-1)^1}{1!}+\frac{20(x-1)^2}{2!}\)

which simplified becomes

\(\displaystyle 11+21(x-1)+10(x-1)^2\)

The Taylor series about x=a for a function is given by

\(\displaystyle \sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}\)

For the first three terms (n=0, 1, 2) we must find the zeroth, first, and second derivative of the function. The zeroth derivative is just the function itself.

\(\displaystyle f'(x)=2x+e^x\)

\(\displaystyle f''(x)=2+e^x\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}\)

Now, using the above formula, write out the first three terms:

\(\displaystyle \frac{(1+e^1)(x-1)^0}{0!}+\frac{[(2+e^1)(x-1)^1]}{1!}+\frac{(2+e^1)(x-1)^2}{2!}\)

which simplified becomes

\(\displaystyle (1+e^1)+[(2+e^1)(x-1)]+\frac{(2+e^1)(x-1)^2}{2}\)

The Taylor series about x=a for a function is

\(\displaystyle \sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}\)

For the first three terms (n=0, 1, 2), we must find the zeroth, first, and second derivative, where the zeroth derivative is just the function itself:

\(\displaystyle f^0=3cos(x)\)

\(\displaystyle f'(x)=-3\sin(x)\)

\(\displaystyle f''(x)=-3\cos(x)\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)\)

Now, using the above formula, write out the first three terms:

\(\displaystyle \frac{3(x-\pi)^0}{0!}-3\cdot \frac{0(x-\pi)^1}{1!}-\frac{3(x-\pi)^2}{2!}\)

which simplified becomes

\(\displaystyle 3+0-3\frac{(x-\pi)^2}{2}\).

The Taylor series about x=a for a function is given by

\(\displaystyle \sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}\)

Now, for the first two terms (n=0, 1) we must find the zeroth and first derivative of the function, the zeroth derivative being the function itself:

\(\displaystyle f'(x)=-\sin^2(x)+\cos^2(x)\)

The derivative was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\)

Now, use the above formula to write out the first two terms:

\(\displaystyle \frac{0(x-\pi)^0}{0!}+\frac{1(x-\pi)^1}{1!}\)

which simplified becomes

\(\displaystyle 0+(x-\pi)\)

The Taylor series about x=a for a function is

\(\displaystyle \sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}\)

For the first four terms (n=0, 1, 2, 3), we must find the zeroth, first, second, and third derivative of the function. The zeroth derivative is the function itself.

\(\displaystyle f'(x)=3x^2+6x+5\)

\(\displaystyle f''(x)=6x+6\)

\(\displaystyle f'''(x)=6\)

The derivatives were found using the following rule:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

Now, use the above formula to write out the first four terms:

\(\displaystyle \frac{10(x-1)^0}{0!}+\frac{14(x-1)^1}{1!}+\frac{12(x-1)^2}{2!}+\frac{6(x-1)^3}{3!}\)

which simplified becomes

\(\displaystyle 10+14(x-1)+6(x-1)^2+(x-1)^3\).

The Taylor series about x=a for a given function is

\(\displaystyle \sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}\)

So, for the first three terms (n=0, 1, 2), we must find the zeroth, first, and second derivatives of the function, where the zeroth derivative is just the function itself:

\(\displaystyle f'(x)=2\sin(x)\cos(x)\)

\(\displaystyle f''(x)=-2\sin^2(x)+2\cos^2(x)\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\) , 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'\(x)\)

Now, using the above formula, write out the first three terms:

\(\displaystyle \frac{0(x-4\pi)^0}{0!}+\frac{0(x-4\pi)^1}{1!}+\frac{2(x-4\pi)^2}{2!}\)

which simplified becomes

\(\displaystyle 0+0+(x-4\pi)^2\).

The Taylor series about x=a for a funtion is given by

\(\displaystyle \sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}\)

For the first four terms (n=0, 1, 2, 3), we must find the zeroth, first, second, and third derivative of the function, where the zeroth derivative is just the function itself:

\(\displaystyle f'(x)=-2\sin(x)+9x^2\)

\(\displaystyle f''(x)-2\cos(x)+18x\)

\(\displaystyle f'''(x)=2\sin(x)+18\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)\)

Now, use the above formula to write out the first four terms (after some distribution we get the final answer):

\(\displaystyle \frac{3(\frac{\pi}{2})^3(x-\frac{\pi}{2})^0}{0!}+\frac{(-2+9(\frac{\pi}{2})^2)(x-\frac{\pi}{2})^1}{1!}+\frac{18(\frac{\pi}{2})(x-\frac{\pi}{2})^2}{2!}+\frac{20(x-\frac{\pi}{2})^3}{3!}\)

which simplified becomes

\(\displaystyle \frac{3\pi^3}{8}+(-2+\frac{9\pi^2}{4})(x-\frac{\pi}{2})+\frac{(9\pi)(x-\frac{\pi}{2})^2}{2}+\frac{10(x-\frac{\pi}{2})^3}{3}\)

The Taylor series about x=a for a function is given by

\(\displaystyle \sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}\)

First, we must find the zeroth and first derivative of the function (n=0, 1), where the zeroth derivative is just the function itself:

\(\displaystyle f'(x)=e^{x-\pi}-3\cos^2(x)\sin(x)\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x)) \cdot g'(x)\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\)

Now, use the above formula to write out the first two terms:

\(\displaystyle \frac{(1-1)(x-\pi)^0}{0!}+\frac{1(x-\pi)^1}{1!}\)

which simplified becomes

\(\displaystyle 0+(x-\pi)\)

The Taylor series about x=a for a function is given by

\(\displaystyle \sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}\)

For the first three terms (n=0, 1, 2) we must find the zeroth, first, and second derivative of the function, where the zeroth derivative is just the function itself:

\(\displaystyle f'(x)=\cos(x)+\sec^2(x)\)

\(\displaystyle f''(x)-\sin(x)+2\sec^2(x)\tan(x)\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)\), 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x)) \cdot g'(x)\)

Now, write out the first three terms using the above formula:

\(\displaystyle \frac{0(x-\pi)^0}{0!}+\frac{0(x-\pi)^1}{1!}+\frac{0(x-\pi)^2}{2!}\)

which simplified becomes

\(\displaystyle 0+0+0\)

It is known that the power series for \(\displaystyle \sin x\) at \(\displaystyle a=0\) is

\(\displaystyle \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}\)

So we just need to plug this in with \(\displaystyle 5x\) instead of \(\displaystyle x\) to get

\(\displaystyle \sin 5x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}(5x)^{2n+1}=\sum_{n=0}^{\infty} \frac{(-1)^n5^{2n+1}}{(2n+1)!}x^{2n+1}\)

\(\displaystyle =5\sum_{n=0}^{\infty} \frac{(-1)^n5^{2n}}{(2n+1)!}x^{2n+1}\)