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Calculus 1 : Differential Functions

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Example Questions

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:

$\frac{4}{9}$

$-\frac{4}{9}$

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

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

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

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 #1861 : Calculus

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:

$\frac{12}{\pi}$

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

$\frac{\pi}{12}$

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

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

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

Example Question #651 : Other Differential Functions

f(x)= (e^x)^2

Find the derivative.

Possible Answers:

$f'(x)= 2e^{x}$

$f'(x)= 2e^{2x}$

f'(x)= 0

$f'(x)= e^{x}$

$f'(x)= e^{2x}$

Correct answer:

$f'(x)= 2e^{2x}$

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

If f(x) = C (constant), then the derivative is f '(x) = 0 .

If f(x) = x^C , then the derivative is $f'(x) = (C) x^ {(C-1)}$ .

If f(x)= ln(x) , then the derivative is $f'(x)= \frac{1}{x}$ .

If f(x)= e^x , the the derivative is f'(x)= e^x .

If $f(x)= \sin x$ , then the derivative is $f'(x)= \cos x$ .

There are many other rules for the derivatives for trig functions.

If F(x)= (f*g)(x) , then the derivative is F'(x)= f'(g(x)) g'(x) . This is known as the chain rule.

In this case, we must find the derivative of the following: f(x)= (e^x)^2

That is done by doing the following: $f'(x)= (2)(e^x)^{(2-1)}(e^x)$

Therefore, the answer is: $f'(x)= 2e^{2x}$

Report an Error

Example Question #651 : Other Differential Functions

$f(x)= \sin x + (x-2)^2$

Find the derivative.

Possible Answers:

$f'(x)= \cos x - 2(x-2)$

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

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

$f'(x)= \cos x + 2(x-2)^2$

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

Correct answer:

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

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

If f(x) = C (constant), then the derivative is f '(x) = 0 .

If f(x) = x^C , then the derivative is $f'(x) = (C) x^ {(C-1)}$ .

If f(x)= ln(x) , then the derivative is $f'(x)= \frac{1}{x}$ .

If f(x)= e^x , the the derivative is f'(x)= e^x .

If $f(x)= \sin x$ , then the derivative is $f'(x)= \cos x$ .

There are many other rules for the derivatives for trig functions.

If F(x)= (f*g)(x) , then the derivative is F'(x)= f'(g(x)) g'(x) . This is known as the chain rule.

In this case, we must find the derivative of the following: $f(x)= \sin x + (x-2)^2$

That is done by doing the following: $f'(x)= \cos x + (2)(x-2)^{(2-1)}(1)$

Therefore, the answer is: $f'(x)= \cos x + 2(x-2)$

Report an Error

Example Question #840 : Functions

$f(x)= x^2 + \frac{1}{6}x^6$

Find the derivative.

Possible Answers:

f'(x) = x^5

f'(x) = 2x + x^5

f'(x) = 2x + x^6

f'(x) = 2x + x^5 -1

f'(x) = 2x

Correct answer:

f'(x) = 2x + x^5

Explanation:

In order to find the derivative of a given function, there are sets of rules you must follow:

If f(x) = C (constant), then the derivative is f '(x) = 0 .

If f(x) = x^C , then the derivative is $f'(x) = (C) x^ {(C-1)}$ .

If f(x)= ln(x) , then the derivative is $f'(x)= \frac{1}{x}$ .

If f(x)= e^x , the the derivative is f'(x)= e^x .

If $f(x)= \sin x$ , then the derivative is $f'(x)= \cos x$ .

There are many other rules for the derivatives for trig functions.

In this case, we must find the derivative of the following: $f(x)= x^2 + \frac{1}{6}x^6$

That is done by doing the following: $f(x)= (2)x^{(2-1)} + (6)\frac{1}{6}x^{(6-1)}$

Therefore, the answer is: f'(x) = 2x + x^5

Report an Error

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

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 #1861 : Calculus

Example Question #651 : Other Differential Functions

Example Question #651 : Other Differential Functions

Example Question #840 : Functions

All Calculus 1 Resources

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