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Calculus 1 : Differential Functions

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Example Questions

Example Question #411 : Other Differential Functions

Find the slope of the line tangent to the function $f(x)=e^{2^{x}}$ at x=0.2 .

Possible Answers:

11.123

4.859

2.511

1.233

0.697

Correct answer:

2.511

Explanation:

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

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

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

Taking the derivative of the function $f(x)=e^{2^{x}}$ at x=0.2 .

The slope of the tangent is

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

$f'(0.2)=2^{0.2}e^{2^{0.2}}ln(2)=2.511$

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Example Question #601 : Differential Functions

Find the slope of the line tangent to the function $f(x)=x+1+\frac{1}{x}$ at x=1 .

Possible Answers:

Correct answer:

Explanation:

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

The slope of the tangent is

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

$f'(1)=1-\frac{1}{1^2}=0$

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Example Question #602 : Functions

Find the slope of the line tangent to the function $f(x)=8^{cos(2x)}$ at $x=\frac{\pi}{4}$ .

Possible Answers:

2ln(8)

-2ln(8)

$\frac{2}{ln(8)}$

$\frac{1}{2ln(8)}$

Correct answer:

-2ln(8)

Explanation:

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)

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 $f(x)=8^{cos(2x)}$ at $x=\frac{\pi}{4}$ .

The slope of the tangent is

$f'(x)=-2sin(2x)(8^{cos(2x)})ln(8)$

$f'(\frac{\pi}{4})=-2sin(2(\frac{\pi}{4}))(8^{cos(2(\frac{\pi}{4}))})ln(8)=-2ln(8)$

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Example Question #411 : Other Differential Functions

Find the slope of the line tangent to the function $f(x)=cos(sin(3^{x}))$ at x=0 .

Possible Answers:

-0.108

1.113

0.715

-0.443

Correct answer:

-0.443

Explanation:

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)

Trigonometric derivative:

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

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

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

Taking the derivative of the function $f(x)=cos(sin(3^{x}))$ at x=0 .

The slope of the tangent is

$f'(x)=-3^{x}cos(3^{x})sin(sin(3^{x}))ln(3)$

$f'(0)=-3^{0}cos(3^{0})sin(sin(3^{0}))ln(3)=-0.443$

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Example Question #603 : Functions

Find the slope of the line tangent to the function $f(x)=4^{x+2}$ at the point x=2 .

Possible Answers:

256ln(4)

64ln(4)

256

Correct answer:

256ln(4)

Explanation:

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

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

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

The slope of the tangent is

$f'(x)=4^{x+2}ln(4)$

$f'(2)=4^{2+2}ln(4)=256ln(4)$

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Example Question #411 : How To Find Differential Functions

Find the slope of the line tangent to the function f(x)=(x^2+x+3)^2 at x=2 .

Possible Answers:

Correct answer:

Explanation:

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.

$d(u^n)=nu^{n-1}du; n:constant$

Note that u may be a complex function, like shown in this problem.

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

The slope of the tangent is

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

f'(2)=2(2(2)+1)(2^2+2+3)=90

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Example Question #607 : Functions

Let $f(x)=x^{\ln(3)}$ on the interval (1,2) . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

1.468

1.992

1.110

1.232

1.581

Correct answer:

1.468

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 $f(x)=x^{\ln(3)}$ on the interval (1,2)

$f(1)=1^{\ln(3)}=1$

$f(2)=2^{\ln(3)}=2.141$

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

$\frac{2.141-1}{2-1}=1.141$

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

$f'(x)=x^{\ln(3)-1}\ln(3)$

$x^{\ln(3)-1}\ln(3)=1.141$

x=1.468 , which verifies the MVT, as it falls within (1,2)

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Example Question #421 : Other Differential Functions

Let $f(x)=\ln^x(3)$ on the interval (0,1) . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

0.147

0.909

0.003

0.282

0.504

Correct answer:

0.504

Explanation:

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 $f(x)=\ln^x(3)$ on the interval (0,1)

$f(0)=\ln^0(3)=1$

$f(1)=\ln^1(3)=\ln(3)$

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

$\frac{\ln(3)-1}{1-0}=\ln(3)-1$

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

$d(a^u)=a^u\ln(a)du$

$f'(x)=\ln^{x}(3)\ln(\ln(3))$

$\ln^{x}(3)\ln(\ln(3))=\ln(3)-1$

$\ln^x(3)=\frac{\ln(3)-1}{\ln(\ln(3))}$

$x=\log_{\ln(3)}(\frac{\ln(3)-1}{\ln(\ln(3))})$

x=0.504 which satisfies the MVT by falling within (0,1)

It may seem daunting, but just treat $\ln(3)$ as another constant.

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Example Question #604 : Differential Functions

Find the slope of the line normal to the function $f(x)=1.5^{x^{1.5}}$ at x=1.5

Possible Answers:

-0.637

0.637

2.218

-1.569

1.569

Correct answer:

-0.637

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:

$d[a^u]=a^udu\ln(a)$

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

Taking the derivative of the function $f(x)=1.5^{x^{1.5}}$ at x=1.5

The slope of the tangent is

$f'(x)=1.5x^{0.5}(1.5^{x^{1.5}})\ln(1.5)$

$f'(x)=1.5(1.5)^{0.5}(1.5^{1.5^{1.5}})\ln(1.5)=1.569$

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

-0.637

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Example Question #605 : Differential Functions

Find the slope of the line normal to the function $f(x)=(1+\ln(4))^{2x}$ at x=3

Possible Answers:

-0.003

321.192

121.008

23.894

-0.118

Correct answer:

-0.003

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:

$d[a^u]=a^udu\ln(a)$

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

Taking the derivative of the function $f(x)=(1+\ln(4))^{2x}$ at x=3

The slope of the tangent is

$f'(x)=2(1+\ln(4))^{2x}\ln(1+\ln(4))$

$f'(3)=2(1+\ln(4))^{2(3)}\ln(1+\ln(4))=321.192$

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

-0.003

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #411 : Other Differential Functions

Example Question #601 : Differential Functions

Example Question #602 : Functions

Example Question #411 : Other Differential Functions

Example Question #603 : Functions

Example Question #411 : How To Find Differential Functions

Example Question #607 : Functions

Example Question #421 : Other Differential Functions

Example Question #604 : Differential Functions

Example Question #605 : Differential Functions

All Calculus 1 Resources

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