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AP Calculus AB : Derivative at a point

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Example Question #41 : Slope Of A Curve At A Point

Find the equation of the line passing through (-1, 1) and perpendicular to the line tangent to the following function at x=4:

f(x)=x^3

Possible Answers:

$y=\frac{x}{48}+\frac{47}{48}$

$y=-\frac{x}{48}+\frac{47}{48}$

y=-{48}x+49

y={48}x+49

Correct answer:

$y=-\frac{x}{48}+\frac{47}{48}$

Explanation:

To determine the equation of the line, we must first find its slope, which is perpendicular to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

f'(x)=3x^2

The derivative was found using the following rule:

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

Evaluating the derivative at the given point, we find that the slope of the tangent line to the function is

f'(4)=48

However, the negative reciprocal of this - the slope of the tangent line - is

$-\frac{1}{48}$

We now can write the equation of the line using point-slope form:

$y-1=-\frac{1}{48}(x+1)$

$y=-\frac{x}{48}+\frac{47}{48}$

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Example Question #61 : Derivative At A Point

Find the equation of the line passing through (4, 10) and parallel to the tangent line to the following function at x=2:

$f=\ln(x^2-x)$

Possible Answers:

y=x

$y=\frac{3}{2}x+4$

$y=\frac{3}{2}x$

$y=-\frac{2}{3}x+16$

Correct answer:

$y=\frac{3}{2}x+4$

Explanation:

To determine the equation of the line, we must first find its slope, which is parallel to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

$f'=\frac{2x-1}{x^2-x}$

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\cdot \frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Now, we find the slope of the tangent line at the given point by plugging this point into the first derivative function:

$f'(2)=\frac{3}{2}$

Now, we use point-slope form to find the equation of the line:

$y-10=\frac{3}{2}(x-4)$

$y=\frac{3}{2}x+4$

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Example Question #43 : Slope Of A Curve At A Point

Find the equation of the line passing through (3,1) and parallel to the tangent line to the following function at x=2:

f=x^3-1

Possible Answers:

y=27

y=12x-2

None of the other answers

y=x

y=12x-35

Correct answer:

y=12x-35

Explanation:

f'=3x^2

The following rule was used to find the derivative:

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

Evaluating the derivative at the given point, we find that the slope of the tangent line to the function is

3(2^2)=12=m

We now can write the equation of the line using point-slope form:

y-1=12(x-3)

y=12x-35

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Example Question #44 : Slope Of A Curve At A Point

Find the equation of the line that passes through (2,0) and is perpendicular to the tangent line to the following function at x=0:

$f=\ln(2-x)$

Possible Answers:

$y=-\frac{1}{2}x+1$

y=2x-4

y=-2x+4

$y=-\frac{1}{2}x+3$

Correct answer:

y=2x-4

Explanation:

To determine the equation of the line, we must first find its slope, which is perpendicular to that of the tangent line to the function. The tangent line to the function at any point is given by the first derivative of the function:

$f'=\frac{-1}{2-x}$

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Evaluating the derivative at the given point, we find that the slope of the tangent line to the function is

$-\frac{1}{2-0}=-\frac{1}{2}$

However, the line we want has a slope perpendicular to this, so we take the negative reciprocal:

m=2

We now can write the equation of the line using point-slope form:

y-0=2(x-2)

y=2x-4

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Example Question #45 : Slope Of A Curve At A Point

Find the slope of the line tangent to the curve of f(x) when x=-2 . Round to the nearest whole number.

f(x)=16e^x+12x^6-48x+13

Possible Answers:

Cannot be determined from the information provided.

2304

Correct answer:

Explanation:

Find the slope of the line tangent to the curve of f(x) when x=-2

f(x)=16e^x+12x^6-48x+13

To find the slope of a tangent line, we need to find our first derivative.

To find our derivative, we need to recall two rules.

$\frac{dy}{dx}e^x=e^x$

And

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

Using these two rules, we can find the derivative of f(x).

Our first term can be derived using our first rule. The derivative of e to the x is just e to the x.

This means that our first term will remain 16e to x.

For our other three terms, we follow the second rule. We will decrease each term's exponent by 1, and then multiply the coefficient by the old exponent.

$f'(x)=16e^x+(6)12x^{6-1}-(1)48x^{1-1}$

Notice that the 13 will drop out. It is a constant term, and as such when we multiply it by it's original exponent (0) it wil be reduced to zero as well.

Clean up the above to get:

$f'(x)=16e^x+72x^{5}-48$

Now, we are almost there. We need to find the slope when x=-2. To do so, plug in -2 for x and solve.

$f'(x)=16e^{-2}+72(-2)^{5}-48\approx -2349.83$

So, our answer is -2350

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Example Question #46 : Slope Of A Curve At A Point

Find the equation of the line passing through the origin and perpendicular to the tangent to the following function at $x=\frac{\pi}{12}$ :

$f=\frac{\cos(3x)}{3}$

Possible Answers:

y=x

$y=\sqrt{2}x$

$y=-\sqrt{2}x$

$y=-\frac{\sqrt{2}x}{2}$

Correct answer:

$y=\sqrt{2}x$

Explanation:

$f'=\frac{-3\sin(3x)}{3}=-\sin(3x)$

The first derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Evaluating the derivative at the given point, we get

$-\sin(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$

Because the slope of the line we want is perpendicular to this line, its slope is the negative reciprocal, or

$\frac{2}{\sqrt{2}}=\sqrt{2}=m$

We can now find the equation of the line using point-slope form:

$y-0=\sqrt{2}(x-0)$

$y=\sqrt{2}x$

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Example Question #51 : Slope Of A Curve At A Point

Find the slope, , of f(x) at x=2 ?

f(x)=2x^5-x^3+4

Possible Answers:

m=164

m=0

m=148

m=128

m=136

Correct answer:

m=148

Explanation:

The slope of any point on this function can be determined by plugging the point's x-value into the f'(x) , the first derivative of f(x) .

f(x)=2x^5-x^3+4

f'(x)=10x^4-3x^2

f'(2)=10(2)^4-3(2)^2

=10(16)-3(4)=160-12=148

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Example Question #52 : Slope Of A Curve At A Point

Find m in y = mx + b from the equation, given the point (2,0)

y = (x^2+2x)(-x^3+4)

Possible Answers:

= 144

= 48

$= -120$

= -96

= 100

Correct answer:

$= -120$

Explanation:

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.

f(x)g'(x)+g(x)f'(x)

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

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

f(x)= x^2 + 2x ; f'(x)= 2x + 2

g(x)= -x^3 +4 ; g'(x) = -3x^2

Use power rule and we get :

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

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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Example Question #821 : Ap Calculus Ab

Which of the following functions contains a removeable discontinuity?

Possible Answers:

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

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

$f(x)=1-(\ln x)^{2}$

$f(x)=\frac{x}{1-x^{2}}$

$f(x)=x^{2}\sin x^{2}$

Correct answer:

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

Explanation:

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at x=a , then the limit as approaches exists, but the value of f(a) does not.

For example, the function $f(x)=\frac{1+x^3}{1+x}$ contains a removeable discontinuity at x=-1 . Notice that we could simplify f(x) as follows:

$f(x)=\frac{1+x^3}{1+x}=\frac{(1+x)(x^2-x+1)}{1+x}=x^{2}-x+1$ , where $x\neq -1$ .

Thus, we could say that $\lim_{x\rightarrow -1}\frac{1+x^3}{1+x}=\lim_{x\rightarrow -1}x^2-x+1=(-1)^2-(-1)+1=3$ .

As we can see, the limit of f(x) exists at x=-1 , even though f(-1) is undefined.

What this means is that f(x) will look just like the parabola with the equation $x^{2}-x+1$ EXCEPT when x=-1 , where there will be a hole in the graph. However, if we were to just define f(-1)=3 , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at x=-1 .

The functions

$f(x)=\frac{x}{1-x^{2}}$ , and

$f(x)=1-(\ln x)^{2}$

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

$f(x)=x^{2}\sin x^{2}$ and $f(x)=\frac{x+1}{1+x^{2}}$ are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

$f(x)=\frac{1+x^{3}}{1+x}$ .

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Example Question #1 : Approximate Rate Of Change From Graphs And Tables Of Values

$\begin{align*}&\text{Approximate a forward derivative at }x=4\\&\text{Using the following table:}\\&x&&f(x)\\&1&&5\\&4&&-8\\&7&&15\end{align*}$

Possible Answers:

$-\frac{13}{3}$

$-\frac{8}{3}$

$\frac{23}{3}$

$\frac{23}{6}$

Correct answer:

$\frac{23}{3}$

Explanation:

$\begin{align*}&\text{To find a forward derivative approximation, we consider function}\\&\text{two values: one at the point of interest and another succeeding}\\&\text{it, and finally that distance between the two. Mathematically,}\\&\text{this follows the form:}\\&f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{Where h is the size of the increment between points on the x-axis.}\\&\text{In approximating a derivative, it is generally prudent to minimize}\\&\text{the distance between function values. In other words, it’s a}\\&\text{good idea to look at two adjacent points. For our problem, that’d}\\&\text{be:}\\&f(4)\text{ and }f(7)\\&\text{Applying the formula above, with }h=3\text{, we find:}\\&f'(4)\approx\frac{15-(15)}{3}\\&f'(4)\approx\frac{23}{3}\end{align*}$

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AP Calculus AB : Derivative at a point

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #41 : Slope Of A Curve At A Point

Example Question #61 : Derivative At A Point

Example Question #43 : Slope Of A Curve At A Point

Example Question #44 : Slope Of A Curve At A Point

Example Question #45 : Slope Of A Curve At A Point

Example Question #46 : Slope Of A Curve At A Point

Example Question #51 : Slope Of A Curve At A Point

Example Question #52 : Slope Of A Curve At A Point

Example Question #821 : Ap Calculus Ab

Example Question #1 : Approximate Rate Of Change From Graphs And Tables Of Values

All AP Calculus AB Resources

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