Slope of a curve at a point - AP Calculus AB

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Question

Find the slope of the line tangent to the curve of d(t) when t is equal to 0.

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Answer

Find the slope of the line tangent to the curve of d(t) when t is equal to 0.

So, we are asked to find the slope of a tangent line at a particular point. To do so, we need to find the derivative of our function, and then evaluate it at the given value of t.

Let's begin by finding d'(t).

To do so, we need to use the power rule, as well as the rules for sine and cosine.

Power Rule

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

Sine Rule

$\frac{d}{dx}(sin(x))=cos(x)$

Cosine Rule

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

Now, we can use all of these to find the derivative of d(t)

Becomes:

Now, we are almost there, but we need to evaluate this derivative when t=0.

This simplifies quite nicely to:

So, the slope of our tangent line when t=0 is negative 17.

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Question

Find the slope of the line tangent to the function

$g(x)=x^2\cos(x)$

at $x=\pi$

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Answer

The slope of the line tangent to a function is nothing more than the first derivative, which is for this function equal to

$g'(x)=2x\cos(x)-x^2\sin(x)$

found using the following rules:

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

Evaluated at the given point, we find the slope of the tangent line equal to

$g'(\pi)=2\pi(\cos(\pi))-\pi^2(\sin(\pi))=-2\pi$

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Question

If $f(x)=|x-1|+3$ , what is the value of $f'(1)$ ?

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Answer

The function is not differentiable at . So the derivative at does not exist.

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Question

Find the slope of the curve , at .

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Answer

Find the derivative of the curve using the power rule.

In mathematical terms, the power rule states,

$\frac{d}{dx}x^n=nx^{n-1}$

Therefore the derivative is,

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

Next, plug in the x-value to find the slope of the curve at , which gives you a final answer of

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Question

Find the slope of the curve, , at .

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Answer

Expand the binomial and combine like terms to get,

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

$\frac{d}{dx}x^n=nx^{n-1}$

Therefore,

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

Lastly, plug in for , to get the slope at the point.

This gives you your final answer of

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Question

Find the slope of the curve $f(x)=\sin(x)+x^2$ at the point with x-coordinate $x=\pi$ .

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Answer

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

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

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

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Question

Find k'(1) if $k(x)=\frac{3x^2+4x-2}{10}$ .

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Answer

First, find the derivative. You should get $k'(x)=\frac{1}{10}(6x+4)$ .

Next, plug in x=1.

You should get .

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Question

Find the slope of $f(t)=\sqrt{5t+6}$ at t=0.

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Answer

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

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

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

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Question

Find the slope at the point (1,2) of the function

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Answer

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. $n(x^{n-1})$

So, we get

From there we plug in our x value.

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Question

Find m in from the equation, given the point (2,0)

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Answer

To find the tangent line at the given point, we need to first take the derivative of the given function.

To find the derivative we need to use product rule. Product rule states that we take the derivative of the first function and multiply it by the derivative of the second function and then add that with the derivative of the second function multiplied by the given first function. To find the derivative of each separate function we need to use 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

$n(x^{n-1})$

Use power rule and we get :

From here, to find the slope at the given point we plug in "2" for x.

$(2^2 + 2(2))(-3(2^2)) + (- (2^3)+4)(2(2)+2)$

This comes out to equal $-120$

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Question

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

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Answer

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

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

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

So, our answer is 1271

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Question

Find the slope of the function $y = xe^{2x}$ at the point $(\ln2,4\ln2)$

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Answer

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

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

Plugging the -value of our point into this equation, we get our desired slope of

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

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Question

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

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Answer

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

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.

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

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

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

$e^x\rightarrow e^x$

The derivative of cosine is negative sine

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

Now, put it all together to get:

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

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

So, our answer is 0

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Question

If , what is the slope of the curve at the point ?

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Answer

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

, so , and plugging in our -value gives .

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Question

$\begin{align}&\text{For positive values of x: }\&f(x)=cos(\frac{(\pi x)}{2}) - 5x + sin(4\cdot \pi x) - 5\&\text{Calculate the slope of the function at }x=6\end{align}$

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Answer

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

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Question

$\begin{align}&\text{Calculate the slope of the function:}\&f(x)=4x + cos(8\cdot \pi x) - sin(\frac{(13\cdot \pi x)}{2}) + 2x^{2} + 3\&\text{At the point }x=2\end{align}$

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Answer

$\begin{align}&\text{Recall the definition of slope as rise over run. To put it in}\&\text{other words, it is a change in height in relation to a change}\&\text{horizontal position: it is a rate. And since it is a rate, it}\&\text{can be found with a derivative. In this case, the derivative}\&\text{of our function (at a height at a given point) with respect}\&\text{to x:}\&Slope=\frac{df(x)}{dx}\&\text{Taking the derivative of our function, }4x + cos(8\cdot \pi x) - sin(\frac{(13\cdot \pi x)}{2}) + 2x^{2} + 3\&\text{We find the slope function: }4x -\frac{ (13\cdot \pi cos(\frac{(13\cdot \pi x)}{2}))}{2}- 8\cdot \pi sin(8\cdot \pi x) + 4\&\text{Plugging in our value of x we then get:}\&f'(2)=4(2) -\frac{ (13\cdot \pi cos(\frac{(13\cdot \pi (2))}{2}))}{2}- 8\cdot \pi sin(8\cdot \pi (2)) + 4\&f'(2)=32.42\end{align}$

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Question

$\begin{align}&\text{Calculate the slope of the function:}\&f(x)=cos(\frac{(13\cdot \pi x)}{2}) + 2sin(4\cdot \pi x) + 2\&\text{At the point }x=5\end{align}$

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Answer

$\begin{align}&\text{Recall the definition of slope as rise over run. To put it in}\&\text{other words, it is a change in height in relation to a change}\&\text{horizontal position: it is a rate. And since it is a rate, it}\&\text{can be found with a derivative. In this case, the derivative}\&\text{of our function (at a height at a given point) with respect}\&\text{to x:}\&Slope=\frac{df(x)}{dx}\&\text{Taking the derivative of our function, }cos(\frac{(13\cdot \pi x)}{2}) + 2sin(4\cdot \pi x) + 2\&\text{We find the slope function: }8\cdot \pi cos(4\cdot \pi x) -\frac{ (13\cdot \pi sin(\frac{(13\cdot \pi x)}{2}))}{2}\&\text{Plugging in our value of x we then get:}\&f'(5)=8\cdot \pi cos(4\cdot \pi (5)) -\frac{ (13\cdot \pi sin(\frac{(13\cdot \pi (5))}{2}))}{2}\&f'(5)=4.71\end{align}$

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Question

$\begin{align}&\text{For positive values of x: }\&f(x)=2sin(4\cdot \pi x) - cos(4\cdot \pi x) - 5x - 2x^{2} - 3x^{3} + 3\&\text{Calculate the slope of the function at }x=2\end{align}$

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Answer

$\begin{align}&\text{Recall the definition of slope as rise over run. To put it in}\&\text{other words, it is a change in height in relation to a change}\&\text{horizontal position: it is a rate. And since it is a rate, it}\&\text{can be found with a derivative. In this case, the derivative}\&\text{of our function (at a height at a given point) with respect}\&\text{to x:}\&Slope=\frac{df(x)}{dx}\&\text{Taking the derivative of our function, }2sin(4\cdot \pi x) - cos(4\cdot \pi x) - 5x - 2x^{2} - 3x^{3} + 3\&\text{We find the slope function: }8\cdot \pi cos(4\cdot \pi x) - 4x + 4\cdot \pi sin(4\cdot \pi x) - 9x^{2} - 5\&\text{Plugging in our value of x we then get:}\&f'(2)=8\cdot \pi cos(4\cdot \pi (2)) - 4(2) + 4\cdot \pi sin(4\cdot \pi (2)) - 9(2)^{2} - 5\&f'(2)=-23.87\end{align}$

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Question

$\begin{align}&\text{Find the slope of: }\&f(x)=x + cos(\frac{(3\cdot \pi x)}{2}) + 4cos(\frac{(3\cdot \pi x)}{4}) - sin(7\cdot \pi x) - x^{2} + 3x^{3} - 2\&\text{At }x=2\end{align}$

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Answer

$\begin{align}&\text{Recall the definition of slope as rise over run. To put it in}\&\text{other words, it is a change in height in relation to a change}\&\text{horizontal position: it is a rate. And since it is a rate, it}\&\text{can be found with a derivative. In this case, the derivative}\&\text{of our function (at a height at a given point) with respect}\&\text{to x:}\&Slope=\frac{df(x)}{dx}\&\text{Taking the derivative of our function, }x + cos(\frac{(3\cdot \pi x)}{2}) + 4cos(\frac{(3\cdot \pi x)}{4}) - sin(7\cdot \pi x) - x^{2} + 3x^{3} - 2\&\text{We find the slope function: }9x^{2} - 7\cdot \pi cos(7\cdot \pi x) -\frac{ (3\cdot \pi sin(\frac{(3\cdot \pi x)}{2}))}{2}- 3\cdot \pi sin(\frac{(3\cdot \pi x)}{4}) - 2x + 1\&\text{Plugging in our value of x we then get:}\&f'(2)=9(2)^{2} - 7\cdot \pi cos(7\cdot \pi (2)) -\frac{ (3\cdot \pi sin(\frac{(3\cdot \pi (2))}{2}))}{2}- 3\cdot \pi sin(\frac{(3\cdot \pi (2))}{4}) - 2(2) + 1\&f'(2)=20.43\end{align}$

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Question

$\begin{align}&\text{Find the slope of: }\&f(x)=4x + cos(9\cdot \pi x) - cos(\frac{(11\cdot \pi x)}{2}) + 5x^{2} + x^{3} - 4\&\text{At }x=2\end{align}$

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Answer

$\begin{align}&\text{Recall the definition of slope as rise over run. To put it in}\&\text{other words, it is a change in height in relation to a change}\&\text{horizontal position: it is a rate. And since it is a rate, it}\&\text{can be found with a derivative. In this case, the derivative}\&\text{of our function (at a height at a given point) with respect}\&\text{to x:}\&Slope=\frac{df(x)}{dx}\&\text{Taking the derivative of our function, }4x + cos(9\cdot \pi x) - cos(\frac{(11\cdot \pi x)}{2}) + 5x^{2} + x^{3} - 4\&\text{We find the slope function: }10x - 9\cdot \pi sin(9\cdot \pi x) +\frac{ (11\cdot \pi sin(\frac{(11\cdot \pi x)}{2}))}{2}+ 3x^{2} + 4\&\text{Plugging in our value of x we then get:}\&f'(2)=10(2) - 9\cdot \pi sin(9\cdot \pi (2)) +\frac{ (11\cdot \pi sin(\frac{(11\cdot \pi (2))}{2}))}{2}+ 3(2)^{2} + 4\&f'(2)=36\end{align}$

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