Slope is defined as the first derivative of a given function.

Since, 

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

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to derive 

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

At the point \(\displaystyle (3,0)\), the \(\displaystyle x\)-coordinate is \(\displaystyle 3\).

Thus, the slope is 

\(\displaystyle f'(3)=2(3)+2=6+2=8\).

Slope is defined as the first derivative of a given function.

Since, 

\(\displaystyle f(x)=7x^{2}-6\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to derive 

\(\displaystyle f'(x)=(2)7x^{1-1}-(0)6=14x\).

At the point \(\displaystyle (5,3)\), the \(\displaystyle x\)-coordinate is \(\displaystyle 5\).

Thus, the slope is \(\displaystyle f'(5)=14(5)=70\).

Slope is defined as the first derivative of a given function.

Since \(\displaystyle f(x)=5x^2-x+7\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\)for all \(\displaystyle n\neq 0\) to determine that 

\(\displaystyle f'(x)=(2)5x^{2-1}-(1)x^{1-1}+(0)7=10x-1.\)

Since we're given a point \(\displaystyle (4,5)\), we can use the \(\displaystyle x\)-coordinate \(\displaystyle 4\) to solve for the slope at that point.

Thus,

\(\displaystyle f'(4)=10(4)-1=40-1=39\).

Slope is defined as the first derivative of a given function.

Since \(\displaystyle f(x)=-2x^2+5x-3\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\) for all \(\displaystyle n\neq 0\) to determine that 

\(\displaystyle f'(x)=(2)(-2)x^{2-1}+(1)5x^{1-1}-(0)3=-4x+5\).

Since we're given a point \(\displaystyle (0,1)\), we can use the x-coordinate \(\displaystyle 0\) to solve for the slope at that point.

Thus, 

\(\displaystyle f'(0)=-4(0)+5=5.\)

Slope is defined as the first derivative of a given function.

Since \(\displaystyle f(x)=-x^2+8x-4\),  we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\)for all \(\displaystyle n\neq 0\) to determine that 

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

Since we're given a point \(\displaystyle (-1,0)\), we can use the x-coordinate \(\displaystyle -1\) to solve for the slope at that point.

Thus, 

\(\displaystyle f'(-1)=-2(-1)+8=2+8=10\).

The slope of the tangent line to a function at a point is the value of the derivative at that point. To calculate the derivative in this problem, the product rule is necessary. Recall that the product rule states that:

\(\displaystyle \frac{d}{dx}(m(x)n(x))=m'(x)n(x)+m(x)n'(x)\).

In this example, \(\displaystyle m(x)=e^x \ and\ n(x)=ln(x).\)

Therefore, 

\(\displaystyle m'(x)=e^x \ and\ n'(x)=\frac{1}{x}\), and

\(\displaystyle \frac{d}{dx}(e^xln(x))=e^xln(x)+e^x\cdot\frac{1}{x}\)

At x = 1, this dervative has the value

\(\displaystyle e^1ln(1)+e^1\cdot\frac{1}{1}=0+e=e\).

The derivative of the function is

\(\displaystyle f'(x)=-2\cos(x)\sin(x)+3\)

and was found using the following rules:

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

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

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

where \(\displaystyle f(x)=\cos^2(x), g(x)=x\) in the chain rule.

Plug in 0 in the derivative function to get \(\displaystyle f'(x)=3\)

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

Since we have

\(\displaystyle f(x)=x^{2}-6x+15\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to determine that 

\(\displaystyle f'(x)=(2)x^{2-1}-(1)6x^{1-1}+(0)15=2x-6\). 

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

\(\displaystyle f'(2)=2(2)-6=4-6=-2\).

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

Since we have

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

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to determine that 

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

We also have a point \(\displaystyle (1,2)\) with a \(\displaystyle x\)-coordinate \(\displaystyle 1\), so the slope 

\(\displaystyle f'(1)=8(1)+1=8+1=9\).

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

Since we have

\(\displaystyle f(x)=-x^{2}-5x-7\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to determine that 

\(\displaystyle f'(x)=(2)(-1)x^{2-1}-(1)5x^{1-1}-(0)7=-2x-5\). 

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

\(\displaystyle f'(4)=-2(4)-5=-8-5=-13\).