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Calculus 1 : How to find differential functions

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Example Questions

Example Question #1585 : Calculus

Determine the slope of the line normal to the function f(x)=ln(cos(x)) at the point $x=\frac{\pi}{3}$ .

Possible Answers:

$-\frac{\sqrt{3}}{3}$

$-\sqrt{3}$

$\sqrt{3}$

$\frac{\sqrt{3}}{3}$

$\frac{\sqrt{3}}{\pi}$

Correct answer:

$\frac{\sqrt{3}}{3}$

Explanation:

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of a natural log:

$d[ln(u)]=\frac{du}{u}$

Trigonometric derivative:

d[sin(u)]=cos(u)du

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function f(x)=ln(cos(x)) at the point $x=\frac{\pi}{3}$ .

The slope of the tangent is

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

$f'(\frac{\pi}{3})=\frac{-sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})}=-\sqrt{3}$

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

$\frac{\sqrt{3}}{3}$ .

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Example Question #1586 : Calculus

Find the slope of the line normal to the function f(x)=tan(2^x) at the point x=3 .

Possible Answers:

0.0038

261.9

-0.0038

0.44

-2.27

Correct answer:

-0.0038

Explanation:

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of an exponential:

d[a^u]=a^uduln(a)

$d[tan(u)]=\frac{du}{cos^2(u)}$

Note that u v may represent large functions, and not just individual variables!

Taking the derivative of the function f(x)=tan(2^x) at the point x=3 .

The slope of the tangent is

$f'(x)=\frac{2^{x}ln(2)}{cos^2(2^{x})}$

$f'(3)=\frac{2^{3}ln(2)}{cos^2(2^{3})}=261.9$

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

-0.0038 .

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Example Question #1587 : Calculus

Find the slope of the line normal to the function $f(x)=\frac{(x^2+2)^2}{x^3}$ at x=4 .

Possible Answers:

$-\frac{64}{45}$

$\frac{45}{64}$

$-\frac{45}{64}$

$-\frac{16}{6561}$

$\frac{64}{45}$

Correct answer:

$-\frac{64}{45}$

Explanation:

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Quotient rule: $d[\frac{u}{v}]=\frac{vdu-udv}{v^2}$

Note that u and v may represent large functions, and not just individual variables!

Taking the derivative of the function $f(x)=\frac{(x^2+2)^2}{x^3}$ at x=4 .

The slope of the tangent is

$f'(x)=\frac{x^3(4x)(x^2+2)-3x^2(x^2+2)^2}{x^6}$

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

$f'(4)=\frac{4(4)^2(4^2+2)-3(4^2+2)^2}{4^4}=\frac{45}{64}$

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

$-\frac{64}{45}$ .

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Example Question #371 : How To Find Differential Functions

Find the slope of the line normal to the function f(x)=x^5+3x^4-6x^3+10 at x=2 .

Possible Answers:

104

$-\frac{1}{104}$

$\frac{1}{104}$

Correct answer:

$-\frac{1}{104}$

Explanation:

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

Taking the derivative of the function f(x)=x^5+3x^4-6x^3+10 at x=2 .

The slope of the tangent is

f'(x)=5x^4+12x^3-18x^2

f'(2)=5(2)^4+12(2)^3-18(2)^2=104

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

$-\frac{1}{104}$ .

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Example Question #1591 : Calculus

Find the slope of the line normal to the function f(x)=x^3-6x^2-4x-24 at x=-2 .

Possible Answers:

$-\frac{1}{24}$

$-\frac{1}{32}$

Correct answer:

$-\frac{1}{32}$

Explanation:

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Taking the derivative of the function f(x)=x^3-6x^2-4x-24 at x=-2 .

The slope of the tangent is

f'(x)=3x^2-12x-4

f'(-2)=3(-2)^2-12(-2)-4=32

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

$-\frac{1}{32}$ .

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Example Question #1592 : Calculus

Find the slope of the line normal to the function f(x)=tan(x^2) at $x=\frac{2}{3}\pi$ .

Possible Answers:

0.08

-0.024

-13.08

-0.08

13.08

Correct answer:

-0.024

Explanation:

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

$d[tan(u)]=\frac{du}{cos^2(u)}$

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function f(x)=tan(x^2) at $x=\frac{2}{3}\pi$ .

The slope of the tangent is

$f'(x)=\frac{2x}{cos^2(x^2)}$

$f'(\frac{2}{3}\pi)=\frac{2(\frac{2}{3}\pi)}{cos^2((\frac{2}{3}\pi)^2)}=40.87$

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

-0.024 .

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Example Question #1593 : Calculus

Find the slope of the line normal to the function $f(x)=tan(e^{x^2})$ at point $x=\pi$ .

Possible Answers:

$-\frac{2\pi e^{\pi^2}}{cos^2(e^{\pi^2})}$

$\frac{2\pi e^{\pi^2}}{cos^2(e^{\pi^2})}$

$-\frac{cos^2(e^{\pi^2})}{2\pi e^{\pi^2}}$

$-\frac{\pi}{tan(e^{x^2})}$

$\frac{cos^2(e^{\pi^2})}{2\pi e^{\pi^2}}$

Correct answer:

$-\frac{cos^2(e^{\pi^2})}{2\pi e^{\pi^2}}$

Explanation:

A line that is normal, that is to say perpendicular to a function at any given point will be normal to this slope of the line tangent to the function at that point.

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

d[e^u]=e^udu

$d[tan(u)]=\frac{du}{cos^2(u)}$

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function $f(x)=tan(e^{x^2})$ at point $x=\pi$ .

The slope of the tangent is

$f'(x)=\frac{2xe^{x^2}}{cos^2(e^{x^2})}$

$f'(\pi)=\frac{2(\pi)e^{\pi^2}}{cos^2(e^{\pi^2})}$

Since the slope of the normal line is perpendicular, it is the negative reciprocal of this value

$-\frac{cos^2(e^{\pi^2})}{2\pi e^{\pi^2}}$ .

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Example Question #1594 : Calculus

Which of the following is not a point of infleciton for the function f(x)=3x^5+20x^4-50x^3+25 ?

Possible Answers:

This function does not have any inflection points.

(2,41)

(0,25)

(-5,9400)

(1,-2)

Correct answer:

(2,41)

Explanation:

The points of inflection of a function occur where the second derivative of the funtion is equal to zero.

Find this second derivative by taking the derivative of the function twice:

f(x)=3x^5+20x^4-50x^3+25

f'(x)=15x^4+80x^3-150x^2

f''(x)=60x^3+240x^2-300x

f''(x)=60x(x^2+4x-5)

Set the second derivative to zero and find the values that satisfy the equation:

f''(x)=60x(x+5)(x-1)

x=-5,0,1

To verify that these are points of inflection, take note how f''(x)=60x^3+240x^2-300x changes sign at these points (i.e. crosses the x-axis) on its graph:

Save

Now, plug these values back in to the original function to find the values of the function that match to them:

f(-5)=3(-5)^5+20(-5)^4-50(-5)^3+25=9400

f(0)=3(0)^5+20(0)^4-50(0)^3+25=25

f(1)=3(1)^5+20(1)^4-50(1)^3+25=-2

Thus the points of inflection are

(-5,9400) , (0,25) , (1,-2)

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Example Question #1595 : Calculus

Find the point of inflection for the function f(x)=7x^3+4x^2+x+5 .

Possible Answers:

$\left(-\frac{8}{21},\frac{6367}{1323}\right)$

$\left(\frac{4}{21},\frac{7121}{1323}\right)$

$\left(-\frac{1}{21},\frac{6563}{1323}\right)$

$\left(-\frac{4}{21},\frac{6491}{1323}\right)$

$\left(\frac{1}{21},\frac{6691}{1323}\right)$

Correct answer:

$\left(-\frac{4}{21},\frac{6491}{1323}\right)$

Explanation:

The points of inflection of a function occur where the second derivative of the funtion is equal to zero.

Find this second derivative by taking the derivative of the function twice:

f(x)=7x^3+4x^2+x+5

f'(x)=21x^2+8x+1

f''(x)=42x+8

Set the second derivative to zero and find the values that satisfy the equation:

42x+8=0

$x=-\frac{4}{21}$

To verify this is a point of inflection, notice how f''(x)=42x+8 changes signs and crosses the x-axis at this point:

Save 1

Now, plug ths value back in to the original function to find the value of the function that matches:

$f(-\frac{4}{21})=7(-\frac{4}{21})^3+4(-\frac{4}{21})^2+(-\frac{4}{21})+5=\frac{6491}{1323}$

The point of inflection is

$\left(-\frac{4}{21},\frac{6491}{1323}\right)$

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Example Question #1 : The Mean Value Theorem

Let f(x)=x^3+7x^2+2x+3 on the interval (0,2) . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

1.05

0.45

-0.83

1.72

-5.72

Correct answer:

1.05

Explanation:

The mean value theorem states that for a planar arc passing through a starting and endpoint (a,b);a<b , there exists at a minimum one point, , within the interval (a,b) for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Note that the value of the derivative of a function at a point is the function's slope at that point; i.e. the slope of the tangent at said point.

First, find the two function values of f(x)=x^3+7x^2+2x+3 on the interval (0,2)

f(0)=0^3+7(0)^2+2(0)+3=3

f(2)=2^3+7(2)^2+2(2)+3=43

Then take the difference of the two and divide by the interval.

$\frac{43-3}{2-0}=20$

Now find the derivative of the function; this will be solved for the value(s) found above.

f'(x)=3x^2+14x+2

3x^2+14x+2=20

x=-5.72,1.05

Since the interval is (0,2) , x=1.05 satisfies the MVT.

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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1585 : Calculus

Example Question #1586 : Calculus

Example Question #1587 : Calculus

Example Question #371 : How To Find Differential Functions

Example Question #1591 : Calculus

Example Question #1592 : Calculus

Example Question #1593 : Calculus

Example Question #1594 : Calculus

Example Question #1595 : Calculus

Example Question #1 : The Mean Value Theorem

All Calculus 1 Resources

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