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Calculus 1 : How to find differential functions

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Example Questions

Example Question #481 : Differential Functions

What is the slope of the line normal to the function f(x)=(x^2+3)^3 at the point x=2 ?

Possible Answers:

$-\frac{2}{343}$

$\frac{1}{588}$

$-\frac{1}{588}$

588

Correct answer:

$-\frac{1}{588}$

Explanation:

The first step to finding the slope of the line normal to a point is to find the slope of the tangent at this point.

The slope of this tangent, in turn, is found by finding the value of the derivative of the function at this point.

Evaluating the function f(x)=(x^2+3)^3 at the point x=2

The slope of the tangent is

f'(x)=3(2x)(x^2+3)^2;f'(2)=3(2(2))(2^2+3)^2=588

The slope of the normal line is the negative reciprocal of this value. Thus for this problem the normal is

$-\frac{1}{588}$

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Example Question #482 : Differential Functions

What is the slope of the function $f(x)=2^{3x}$ at the point x=3 ?

Possible Answers:

-1536ln(2)

$-\frac{1}{1536ln(2)}$

$-\frac{ln(2)}{1536}$

$\frac{ln(2)}{1536}$

$\frac{1536}{ln(2)}$

Correct answer:

$-\frac{1}{1536ln(2)}$

Explanation:

The first step to finding the slope of the line normal to a point is to find the slope of the tangent at this point.

The slope of this tangent, in turn, is found by finding the value of the derivative of the function at this point.

Note that the derivation of an exponential follows the form d[a^u]=a^uduln(a)

Evaluating the function $f(x)=2^{3x}$ at the point x=3

The slope of the tangent is

$f'(x)=3(2^{3x})ln(2);f'(3)=3(2^{3(3)})ln(2)=1536ln(2)$

The slope of the normal line is the negative reciprocal of this value. Thus for this problem the normal is

$-\frac{1}{1536ln(2)}$

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Example Question #483 : Differential Functions

What is the slope of the line normal to the function f(x)=-x^3-x^2+7x+6 at the point x=3 ?

Possible Answers:

$\frac{1}{26}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{26}$

Explanation:

The first step to finding the slope of the line normal to a point is to find the slope of the tangent at this point.

The slope of this tangent, in turn, is found by finding the value of the derivative of the function at this point.

Evaluating the function f(x)=-x^3-x^2+7x+6 at the point x=3

The slope of the tangent is

f'(x)=-3x^2-2x+7;f'(3)=-3(3)^2-2(3)+7=-26

The slope of the normal line is the negative reciprocal of this value. Thus for this problem the normal is

$\frac{1}{26}$

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Example Question #491 : Differential Functions

What is the slope of the line normal to the function f(x)=x^2(x+2)^2 at the point x = 3 ?

Possible Answers:

$-\frac{1}{15}$

$-\frac{1}{240}$

240

Correct answer:

$-\frac{1}{240}$

Explanation:

The first step to finding the slope of the line normal to a point is to find the slope of the tangent at this point.

The slope of this tangent, in turn, is found by finding the value of the derivative of the function at this point.

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

Product rule: d[uv]=udv+vdu

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

Evaluating the function f(x)=x^2(x+2)^2 at the point x = 3

The slope of the tangent is

f'(x)=2x(x+2)^2+2x^2(x+2);f'(3)=2(3)(3+2)^2+2(3)^2(3+2)=240

The slope of the normal line is the negative reciprocal of this value. Thus for this problem the normal is

$-\frac{1}{240}$

Report an Error

Example Question #492 : Differential Functions

What is the slope of the line normal to the function f(x)=x^5+x^4+x^3+x^2+x+1 at the point x=1 ?

Possible Answers:

$-\frac{1}{15}$

$\frac{1}{15}$

$-\frac{1}{6}$

Correct answer:

$-\frac{1}{15}$

Explanation:

The first step to finding the slope of the line normal to a point is to find the slope of the tangent at this point.

The slope of this tangent, in turn, is found by finding the value of the derivative of the function at this point.

Evaluating the function f(x)=x^5+x^4+x^3+x^2+x+1 at the point x=1

The slope of the tangent is

f'(x)=5x^4+4x^3+3x^2+2x+1

f'(1)=5(1)^4+4(1)^3+3(1)^2+2(1)+1=15

The slope of the normal line is the negative reciprocal of this value. Thus for this problem the normal is

$-\frac{1}{15}$

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Example Question #493 : Differential Functions

What is the slope of the line normal to the function $f(x)=\frac{1}{(x+2)^2}$ at the point x=4 ?

Possible Answers:

$-\frac{1}{108}$

108

$\frac{1}{144}$

Correct answer:

108

Explanation:

The first step to finding the slope of the line normal to a point is to find the slope of the tangent at this point.

The slope of this tangent, in turn, is found by finding the value of the derivative of the function at this point.

Evaluating the function $f(x)=\frac{1}{(x+2)^2}=(x+2)^{-2}$ at the point x=4

The slope of the tangent is

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

$f'(4)=\frac{-2}{(4+2)^3}=-\frac{1}{108}$

The slope of the normal line is the negative reciprocal of this value. Thus for this problem the normal is

108

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Example Question #494 : Differential Functions

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

Possible Answers:

-0.800

0.800

-1.250

1.250

-0.481

Correct answer:

1.250

Explanation:

The first step to finding the slope of the line normal to a point is to find the slope of the tangent at this point.

The slope of this tangent, in turn, is found by finding the value of the derivative of the function at this point.

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

Derivative of an exponential:

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

Trigonometric derivative:

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

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

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

The slope of the tangent is

$f'(x)=-sin(x)(2^{cos(x)})ln(2)$

$f'(\frac{\pi}{4})=-sin(\frac{\pi}{4})(2^{cos(\frac{\pi}{4})})ln(2)=-0.800$

The slope of the normal line is the negative reciprocal of this value. Thus for this problem the normal is

1.250

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Example Question #495 : Differential Functions

What is the slope of the line normal to the function f(x)=sin(2x)cos(3x) at the point $x=\frac{\pi}{4}$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The first step to finding the slope of the line normal to a point is to find the slope of the tangent at this point.

The slope of this tangent, in turn, is found by finding the value of the derivative of the function at this point.

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

Trigonometric derivative:

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

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!

Evaluating the function f(x)=sin(2x)cos(3x) at the point $x=\frac{\pi}{4}$

The slope of the tangent is

f'(x)=2cos(2x)cos(3x)-3sin(2x)sin(3x)

$f'(\frac{\pi}{4})=2cos(2(\frac{\pi}{4}))cos(3(\frac{\pi}{4}))-3sin(2(\frac{\pi}{4}))sin(3(\frac{\pi}{4}))$

$f'(\frac{\pi}{4})=\frac{-3\sqrt{2}}{2}$

The slope of the normal line is the negative reciprocal of this value. Thus for this problem the normal is

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

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Example Question #496 : Differential Functions

What is the slope of the line normal to the function $f(x)=4^{e^x}$ at the point x=ln(2) ?

Possible Answers:

$\frac{1}{32ln(4)}$

32ln(4)

$\frac{16}{ln(2)}$

$-\frac{1}{32ln(4)}$

$-\frac{ln(2)}{16}$

Correct answer:

$-\frac{1}{32ln(4)}$

Explanation:

The first step to finding the slope of the line normal to a point is to find the slope of the tangent at this point.

The slope of this tangent, in turn, is found by finding the value of the derivative of the function at this point.

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

Derivative of an exponential:

d[e^u]=e^udu

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

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

Evaluating the function $f(x)=4^{e^x}$ at the point x=ln(2)

The slope of the tangent is

$f'(x)=e^x4^{e^x}ln(4)$

$f'(ln(2))=e^{ln(2)}4^{e^{ln(2)}}ln(4)=32ln(4)$

The slope of the normal line is the negative reciprocal of this value. Thus for this problem the normal is

$-\frac{1}{32ln(4)}$

Report an Error

Example Question #301 : How To Find Differential Functions

Find the derivative.

$\frac{1}{x}$

Possible Answers:

$-\frac{1}{x^{2}}$

$\frac{1}{x^{2}}$

$x^{2}$

Correct answer:

$-\frac{1}{x^{2}}$

Explanation:

Use the quotient rule to find this derivative.

$\frac{x(0)-1(1)}{x^{2}}$

$\frac{d}{dx}=-\frac{1}{x^{2}}$

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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 #481 : Differential Functions

Example Question #482 : Differential Functions

Example Question #483 : Differential Functions

Example Question #491 : Differential Functions

Example Question #492 : Differential Functions

Example Question #493 : Differential Functions

Example Question #494 : Differential Functions

Example Question #495 : Differential Functions

Example Question #496 : Differential Functions

Example Question #301 : How To Find Differential Functions

All Calculus 1 Resources

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