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Calculus 1 : How to find differential functions

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Example Questions

Example Question #641 : Other Differential Functions

Determine the slope of the line that is normal to the function $f(x)=\frac{cos^2(x)}{4}$ at the point $x=\frac{\pi}{4}$

Possible Answers:

$-\frac{1}{4}$

$\frac{1}{4}$

Correct answer:

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):

Trigonometric derivative:

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

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

Taking the derivative of the function $f(x)=\frac{cos^2(x)}{4}$ at the point $x=\frac{\pi}{4}$

The slope of the tangent is

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

$f'(\frac{\pi}{4})=\frac{-sin(\frac{\pi}{4})cos(\frac{\pi}{4})}{2}=\frac{-1}{4}$

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

$m=-\frac{1}{\frac{-1}{4}}=4$

Report an Error

Example Question #642 : Other Differential Functions

Determine the slope of the line that is normal to the function $f(x)=xe^{cos(x)}$ at the point $x=\pi$

Possible Answers:

$\frac{e}{2}$

$\frac{2}{e}$

$-\frac{e}{2}$

$-\frac{2}{e}$

Correct answer:

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[e^u]=e^udu

Trigonometric derivative:

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

Product rule: d[uv]=udv+vdu

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

Taking the derivative of the function function $f(x)=xe^{cos(x)}$ at the point $x=\pi$

The slope of the tangent is

$f'(x)=e^{cos(x)}-xsin(x)e^{cos(x)}$

$f'(\pi)=e^{cos(\pi)}-\pi sin(\pi)e^{cos(\pi)}=\frac{1}{e}$

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

$m=-\frac{1}{\frac{1}{e}}=-e$

Report an Error

Example Question #643 : Other Differential Functions

Determine the slope of the line that is normal to the function f(x)=(x^4+x^2-1)^4 at the point x=1

Possible Answers:

$-\frac{1}{24}$

$-\frac{1}{12}$

$\frac{1}{12}$

Correct answer:

$-\frac{1}{24}$

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^4+x^2-1)^4 at the point x=1

The slope of the tangent is

f'(x)=4(4x^3+2x)(x^4+x^2-1)^3

f'(1)=4(4(1)^3+2(1))(1^4+1^2-1)^3=24

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

$m=-\frac{1}{24}$

Report an Error

Example Question #644 : Other Differential Functions

Determine the slope of the line that is normal to the function f(x)=cos^3(2x) at the point $x=\frac{\pi}{12}$

Possible Answers:

$-\frac{3\sqrt{3}}{4}$

$-\frac{4}{9}$

$\frac{4\sqrt{3}}{9}$

$\frac{4}{9}$

$\frac{3\sqrt{3}}{4}$

Correct answer:

$\frac{4}{9}$

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):

Trigonometric derivative:

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

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

Taking the derivative of the function f(x)=cos^3(2x) at the point $x=\frac{\pi}{12}$

The slope of the tangent is

f'(x)=-6sin(2x)cos^2(2x)

$f'(\frac{\pi}{12})=-6sin(2(\frac{\pi}{12}))cos^2(2(\frac{\pi}{12}))=\frac{-9}{4}$

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

$m=-\frac{1}{\frac{-9}{4}}=\frac{4}{9}$

Report an Error

Example Question #831 : Differential Functions

Determine the slope of the line that is normal to the function f(x)=tan^4(3x) at the point $x=\frac{\pi}{4}$

Possible Answers:

$-\frac{1}{36}$

$\frac{1}{36}$

$\frac{1}{24}$

Correct answer:

$\frac{1}{24}$

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):

Trigonometric derivative:

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

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

Taking the derivative of the function f(x)=tan^4(3x) at the point $x=\frac{\pi}{4}$

The slope of the tangent is

$f'(x)=\frac{12tan^3(3x)}{cos^2(3x)}$

$f'(\frac{\pi}{4})=\frac{12tan^3(3(\frac{\pi}{4}))}{cos^2(3(\frac{\pi}{4}))}=-24$

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

$m=-\frac{1}{-24}=\frac{1}{24}$

Report an Error

Example Question #646 : Other Differential Functions

Determine the slope of the line that is normal to the function $f(x)=cos^3(2\pi x^3)$ at the point $x=\frac{1}{2}$

Possible Answers:

$\frac{8\sqrt{2}}{9\pi}$

$-\frac{9\sqrt{2}\pi}{16}$

$\frac{9\sqrt{2}\pi}{16}$

$\frac{-3\sqrt{2}\pi}{4}$

$-\frac{8\sqrt{2}}{9\pi}$

Correct answer:

$\frac{8\sqrt{2}}{9\pi}$

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):

Trigonometric derivative:

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

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

Taking the derivative of the function $f(x)=cos^3(2\pi x^3)$ at the point $x=\frac{1}{2}$

The slope of the tangent is

$f'(x)=-18\pi x^2sin(2\pi x^3)cos^2(2\pi x^3)$

$f'(\frac{1}{2})=-9\pi (\frac{1}{2})^2sin(2\pi (\frac{1}{2})^3)cos^2(2\pi (\frac{1}{2})^3)=\frac{-9\sqrt{2}\pi}{16}$

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

$m=-\frac{1}{\frac{-9\sqrt{2}\pi}{16}}=\frac{8\sqrt{2}}{9\pi}$

Report an Error

Example Question #647 : Other Differential Functions

Determine the slope of the line that is normal to the function f(x)=cos^4(x) at the point $x=\frac{3}{4}\pi$

Possible Answers:

$\frac{1}{4}$

$-\frac{1}{4}$

Correct answer:

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):

Trigonometric derivative:

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

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

Taking the derivative of the function f(x)=cos^4(x) at the point $x=\frac{3}{4}\pi$

The slope of the tangent is

f'(x)=-4sin(x)cos^3(x)

$f'(\frac{3}{4}\pi)=-4sin(\frac{3}{4}\pi)cos^3(\frac{3}{4}\pi)=1$

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

$m=-\frac{1}{1}=-1$

Report an Error

Example Question #648 : Other Differential Functions

Determine the slope of the line that is normal to the function f(x)=sin^4(x) at the point $x=\frac{\pi}{3}$

Possible Answers:

$-\frac{4\sqrt{3}}{3}$

$\frac{3\sqrt{2}}{4}$

$\frac{4\sqrt{3}}{9}$

$\frac{4\sqrt{3}}{3}$

$-\frac{4\sqrt{3}}{9}$

Correct answer:

$-\frac{4\sqrt{3}}{9}$

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):

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)=sin^4(x) at the point $x=\frac{\pi}{3}$

The slope of the tangent is

f'(x)=4cos(x)sin^3(x)

$f'(\frac{\pi}{3})=4cos(\frac{\pi}{3})sin^3(\frac{\pi}{3})=\frac{3\sqrt{3}}{4}$

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

$m=-\frac{1}{\frac{3\sqrt{3}}{4}}=-\frac{4\sqrt{3}}{9}$

Report an Error

Example Question #649 : Other Differential Functions

Determine the slope of the line that is normal to the function f(x)=tan^3(x) at the point $x=\frac{\pi}{3}$

Possible Answers:

$\frac{1}{36}$

$-\frac{\sqrt{3}}{36}$

$12\sqrt{3}$

$\frac{\sqrt{3}}{36}$

$-\frac{1}{36}$

Correct answer:

$-\frac{1}{36}$

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):

Trigonometric derivative:

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

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

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

The slope of the tangent is

$f'(x)=\frac{3tan^2(x)}{cos^2(x)}$

$f'(\frac{\pi}{3})=\frac{3tan^2(\frac{\pi}{3})}{cos^2(\frac{\pi}{3})}=36$

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

$m=-\frac{1}{36}$

Report an Error

Example Question #650 : Other Differential Functions

Determine the slope of the line that is normal to the function $f(x)=tan^4(\frac{\pi}{4}x^2)$ at the point x=3

Possible Answers:

$-\frac{\pi}{12}$

$-\frac{12}{\pi}$

$\frac{\pi}{12}$

$-\frac{1}{12\pi}$

$\frac{12}{\pi}$

Correct answer:

$-\frac{1}{12\pi}$

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):

Trigonometric derivative:

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

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

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

The slope of the tangent is

$f'(x)=\frac{4(\frac{2\pi}{4}x)(tan^3(\frac{\pi}{4}x^2))}{cos^2(\frac{\pi}{4}x^2)}=\frac{2\pi xtan^3(\frac{\pi}{4}x^2)}{cos^2(\frac{\pi}{4}x^2)}$

$f'(3)=\frac{2\pi (3)tan^3(\frac{\pi}{4}(3)^2)}{cos^2(\frac{\pi}{4}(3)^2)}=12\pi$

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

$m=-\frac{1}{12\pi}$

Report an Error

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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 #641 : Other Differential Functions

Example Question #642 : Other Differential Functions

Example Question #643 : Other Differential Functions

Example Question #644 : Other Differential Functions

Example Question #831 : Differential Functions

Example Question #646 : Other Differential Functions

Example Question #647 : Other Differential Functions

Example Question #648 : Other Differential Functions

Example Question #649 : Other Differential Functions

Example Question #650 : Other Differential Functions

All Calculus 1 Resources

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