To find the slope at a point, we take the derivative of \(\displaystyle f(x)\), substitute in \(\displaystyle x=\pi\), and simplify.

\(\displaystyle f'(x)= \cos(x)+2x\).

\(\displaystyle m=f'(\pi)=\cos(\pi)+2\pi = -1+2\pi = 2\pi-1\).

First, use the chain rule to find f'(t).

You should get \(\displaystyle f'(t)=\frac{5}{2}(5t+6)^{-\frac{1}{2}}\).

Next, plug in t=0 to get f'(0)=\(\displaystyle \frac{5}{2\sqrt{6}}\).

Find the derivative of the curve using the power rule.

In mathematical terms, the power rule states,

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

Therefore the derivative is,

 \(\displaystyle \frac{d}{dx}3x^2\rightarrow 2\cdot 3x^{2-1}=6x\) 

Next, plug in the x-value to find the slope of the curve at \(\displaystyle x=4\), which gives you a final answer of \(\displaystyle m=24.\)

Expand the binomial and combine like terms to get, 

\(\displaystyle x^2+12x+39.\) 

Next, take the derivative of the polynomial using the power rule, which is in mathematical terms,

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

Therefore,

\(\displaystyle \frac{d}{dx}x^2+12x+39\rightarrow 2x^{2-1}+1\cdot 12x^{1-1}=2x+12\)

Lastly, plug in \(\displaystyle 0\) for \(\displaystyle x\), to get the slope at the point.

This gives you your final answer of \(\displaystyle m=12.\)

Derivatives are slope finders. Therefore to find the slope at the given point, we need to find the derivative of the function using power rule. Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent.\(\displaystyle n(x^{n-1})\)

So, we get \(\displaystyle y' =3x^2 + 2\)

From there we plug in our x value.

\(\displaystyle 3(1)^2 +2 = 5\)

\(\displaystyle slope=5\)

To find the slope at a point of our function, we need to find its derivative first.

Using the product rule (and the chain rule within this product rule application), we have

\(\displaystyle y' = (x)(2e^{2x})+(e^{2x})(1) = (2x+1)e^{2x}\).

Plugging the \(\displaystyle x\)-value of our point into this equation, we get our desired slope of

\(\displaystyle (2\ln2+1)e^{2\ln2} = (2\ln2+1)e^{\ln4}=(2\ln2+1)4 = 8\ln2 +4\).

To find the slope at a point, we first find the derivative of our function, and then substitute in the \(\displaystyle x\)-value of our point.

\(\displaystyle y = 2x^2+5\), so \(\displaystyle y' = 4x\), and plugging in our \(\displaystyle x\)-value gives \(\displaystyle 4(4)=16\).

Find the rate of change of f(x) at the point (4,12)

\(\displaystyle f(x)=5x^4-3x^2+15x-17\)

We are asked to find a rate of change, so begin by finding the first derivative.

\(\displaystyle f(x)=5x^4-3x^2+15x-17\)

\(\displaystyle f'(x)=20x^3-6x+15\)

Now, we need the rate of change at (4,12). What matters most is the x value, simply plug it into our derivative and solve for y

\(\displaystyle f'(4)=20(4)^3-6(4)+15=1280-24+15=1271\)

So, our answer is 1271

Find the slope of the line tangent to f(x) at the point x=0

\(\displaystyle f(x)=16x^3+e^x+cos(x)\)

To find the slope of a tangent line, we must first find the derivative of our function.

Let's recall a few rules to help us out.

1) The derivative of a monomial can be found by multiplying the coefficient by the exponent, and then decreasing the exponent by 1.

\(\displaystyle 16x^3\rightarrow (3)16x^2=48x^2\)

2) The derivative of e to the x is e to the x

\(\displaystyle e^x\rightarrow e^x\)

3) The derivative of cosine is negative sine

\(\displaystyle cos(x)\rightarrow-sin(x)\)

Now, put it all together to get:

\(\displaystyle f'(x)=48x^2+e^x-sin(x)\)

Lastly, we need to plug in 0 for x and solve our equation.

\(\displaystyle f'(0)=48(0)^2+e^{(0)}-sin(0)=0+1-1=0\)

So, our answer is 0

\(\displaystyle \begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }cos(\frac{(\pi x)}{2}) - 5x + sin(4\cdot \pi x) - 5\\&\text{We find the slope function: }4\cdot \pi cos(4\cdot \pi x) -\frac{ (\pi sin(\frac{(\pi x)}{2}))}{2}- 5\\&\text{Plugging in our value of x we then get:}\\&f'(6)=4\cdot \pi cos(4\cdot \pi (6)) -\frac{ (\pi sin(\frac{(\pi (6))}{2}))}{2}- 5\\&f'(6)=7.57\end{align*}\)