The derivative of the function is

\(\displaystyle f'(x)=20x^3e^{\sin(x)}+5x^4\cos(x)e^{\sin(x)}\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}e^u=\frac{\mathrm{du} }{\mathrm{d} x}e^u\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\sin(x)=\cos(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}\)

To finish, plug in the point given into the first derivative function:

\(\displaystyle f'(0)=0+0=0\)

The derivative of the function is:

\(\displaystyle f'(x)=2x-2\csc(2x)\cot(2x)+1\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\csc(x)=-\csc(x)\cot(x)\), \(\displaystyle e^{\ln(x)}=x\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\)

Now, just plug in the point x=2 into the first derivative function:

\(\displaystyle f'(2)=5-2\csc(4)\cot(4)\)

We define slope as the first derivative of a given function.

Since we have

\(\displaystyle f(x)=2x^{2}-7x+11\), we can use the Power Rule

for all  

to determine that

\(\displaystyle f'(x)=(2)2x^{2-1}-(1)7x^{1-1}+(0)11=4x-7\).

We also have a point  \(\displaystyle (5,6)\) with a \(\displaystyle x\)-coordinate \(\displaystyle 5\), so the slope

\(\displaystyle f'(5)=4(5)-7=20-7=13\).

We define slope as the first derivative of a given function.

Since we have 

\(\displaystyle f(x)=-5x^{2}+\frac{1}{2}x+4\), we can use the Power Rule  

 for all  

to determine that 

\(\displaystyle f'(x)=(2)(-5)x^{2-1}+(1)\frac{1}{2}x^{1-1}+(0)4=-10x+\frac{1}{2}\).

We also have a point  \(\displaystyle (2,5)\) with a \(\displaystyle x\)-coordinate \(\displaystyle 2\), so the slope 

\(\displaystyle f'(2)=-10(2)+\frac{1}{2}=-20+\frac{1}{2}=-\frac{40}{2}+\frac{1}{2}=-\frac{39}{2}\).

We define slope as the first derivative of a given function.

Since we have 

\(\displaystyle f(x)=\frac{1}{2}x^{2}-x+7\), we can use the Power Rule

 for all  to determine that 

\(\displaystyle f'(x)=2(\frac{1}{2}x^{2-1})-(1)x^{1-1}+(0)7=x-1\).

We also have a point  \(\displaystyle (-1,1)\) with a \(\displaystyle x\)-coordinate \(\displaystyle -1\), so the slope 

\(\displaystyle f'(-1)=(-1)-1=-2\).

We define slope as the first derivative of a given function.

Since we have

\(\displaystyle f(x)=-x^{2}+5x-2\), we can use the Power Rule  

for all  

to determine that

\(\displaystyle f'(x)=(2)(-1)x^{2-1}+(1)5x^{1-1}-(0)2=-2x+5\).

We also have a point \(\displaystyle (2,-5)\) with a -\(\displaystyle x\)coordinate \(\displaystyle 2\), so the slope

\(\displaystyle f'(2)=-2(2)+5=-4+5=1\).

We define slope as the first derivative of a given function.

Since we have 

\(\displaystyle f(x)=4x^{2}-3x+2\), we can use the Power Rule

 for all  to determine that 

\(\displaystyle f'(x)=(2)4x^{2-1}-(1)3x^{1-1}+(0)2=8x-3\).

We also have a point \(\displaystyle (1,-1)\) with a \(\displaystyle x\)-coordinate \(\displaystyle 1\), so the slope 

\(\displaystyle f'(1)=8(1)-3=8-3=5\).

We define slope as the first derivative of a given function.

Since we have 

\(\displaystyle f(x)=-x^{2}+11x+4\), we can use the Power Rule  

 for all  

to determine that 

\(\displaystyle f'(x)=(2)(-1)x^{2-1}+(1)11x^{1-1}+(0)4=-2x+11\).

We also have a point \(\displaystyle (4,-5)\) with a \(\displaystyle x\)-coordinate \(\displaystyle 4\), so the slope 

\(\displaystyle f'(4)=-2(4)+11=-8+11=3\).

In order to find \(\displaystyle f'(1)\), we first find \(\displaystyle f'(x)\).

\(\displaystyle f(x)=x^3+x^2+x+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}\)

\(\displaystyle f'(x)=3x^2+2x+1-\frac{1}{x^2}-\frac{2}{x^3}-\frac{3}{x^4}\)

Now we plug in 1 to get

\(\displaystyle f'(1)=3(1)^2+2(1)+1-\frac{1}{(1)^2}-\frac{2}{(1)^3}-\frac{3}{(1)^4}\)

\(\displaystyle f'(1)=3+2+1-1-2-3=0\)

We define slope as the first derivative of a given function.

Since we have \(\displaystyle f(x)=3x^{2}-14\), we can use the Power Rule  

for all  

to determine that

\(\displaystyle f'(x)=(2)3x^{2-1}-(0)14=6x\) .

We also have a point \(\displaystyle (4,-2)\) with a \(\displaystyle x\)-coordinate \(\displaystyle 4\), so the slope

\(\displaystyle f'(4)=6(4)=24\).