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Calculus 1 : Differential Functions

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Example Questions

Example Question #861 : Functions

Determine the slope of the line that is tangent to the function f(x)=ln(ln(cos(x))) at the point x=1.5

Possible Answers:

34.78

-0.38

-5.32

-34.78

5.32

Correct answer:

5.32

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 a natural log:

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

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)=ln(ln(cos(x))) at the point x=1.5

The slope of the tangent is

$f'(x)=\frac{-sin(x)}{cos(x)}\frac{1}{ln(cos(x))}=\frac{-tan(x)}{ln(cos(x))}$

$f'(1.5)=\frac{-tan(1.5)}{ln(cos(1.5))}=5.32$

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Example Question #861 : Functions

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

Possible Answers:

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

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

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

$\sqrt{2}$

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

Correct answer:

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

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

The slope of the tangent is

$f'(x)=\frac{cos(x)}{sin(x)}\frac{1}{ln(sin(x))}=\frac{cot(x)}{ln(sin(x))}$

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

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Example Question #861 : Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=cos(x^2+4x+\frac{\pi}{2})$ at the point x=0

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.

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 $f(x)=cos(x^2+4x+\frac{\pi}{2})$ at the point x=0

The slope of the tangent is

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

$f'(0)=-(2(0)+4)sin(0^2+4(0)+\frac{\pi}{2})=-4$

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Example Question #862 : Functions

Determine the slope of the line that is tangent to the function $f(x)=sin(\pi x+\pi)$ at the point x=3

Possible Answers:

$-\pi$

$3\pi$

$-4\pi$

$4\pi$

$\pi$

Correct answer:

$\pi$

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):

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 $f(x)=sin(\pi x+\pi)$ at the point x=3

The slope of the tangent is

$f'(x)=\pi cos(\pi x+\pi)$

$f'(3)=\pi cos(3\pi +\pi)=\pi$

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Example Question #861 : Differential Functions

Determine the slope of the line that is tangent to the function f(x)=x^6+3x^4+8x^2+10 at the point 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^6+3x^4+8x^2+10 at the point x=1

The slope of the tangent is

f'(x)=6x^5+12x^3+16x

f'(1)=6(1)^5+12(1)^3+16(1)=34

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Example Question #679 : Other Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=cos(\pi(x^2+6x+4))$ at the point x=2

Possible Answers:

$-\frac{7\sqrt{2}}{2}\pi$

$\frac{7\sqrt{2}}{2}\pi$

$-\frac{7}{2}\pi$

$7\sqrt{2}\pi$

$\frac{7}{2}\pi$

Correct answer:

$\frac{7\sqrt{2}}{2}\pi$

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):

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 $f(x)=cos(\pi(x^2+6x+4))$ at the point $x=\frac{1}{2}$

The slope of the tangent is

$f'(x)=-\pi(2x+6)sin(\pi(x^2+6x+4))$

$f'(\frac{1}{2})=-\pi(2(\frac{1}{2})+6)sin(\pi(\frac{1}{2}^2+6(\frac{1}{2})+4))=\frac{7\sqrt{2}}{2}\pi$

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Example Question #680 : Other Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=e^{sin(\pi x^2)}$ at the point x=3

Possible Answers:

$-3\pi$

$-6\pi$

$3\pi$

$6\pi$

Correct answer:

$-6\pi$

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

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 $f(x)=e^{sin(\pi x^2)}$ at the point x=3

The slope of the tangent is

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

$f'(3)=2\pi (3)cos(\pi (3)^2)e^{sin(\pi (3)^2)}=-6\pi$

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Example Question #861 : Functions

Determine the slope of the line that is tangent to the function $f(x)=xe^{x^2}$ at the point x=2

Possible Answers:

2e^4

8e^4

3e^4

9e^4

e^4

Correct answer:

9e^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):

Derivative of an exponential:

d[e^u]=e^udu

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 $f(x)=xe^{x^2}$ at the point x=2

The slope of the tangent is

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

$f'(2)=e^{2^2}+2(2)^2e^{2^2}=9e^4$

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Example Question #682 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function f(x)=sin(xe^x) at the point x=0

Possible Answers:

e+1

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.

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

Derivative of an exponential:

d[e^u]=e^udu

Trigonometric derivative:

d[sin(u)]=cos(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 f(x)=sin(xe^x) at the point x=0

The slope of the tangent is

$f'(x)=(e^x+xe^x)cos(xe^{x})$

f'(0)=(e^0+(0)e^0)cos((0)e^0})=1

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Example Question #683 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=tan(\frac{\pi x^2}{4})$ at the point x=6

Possible Answers:

$3\pi$

$-6\pi$

$-3\pi$

$6\pi$

Correct answer:

$3\pi$

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):

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 $f(x)=tan(\frac{\pi x^2}{4})$ at the point x=6

The slope of the tangent is

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

$f'(6)=\frac{\frac{6}{2}\pi}{cos^2(\frac{\pi (6)^2}{4})}=3\pi$

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #861 : Functions

Example Question #861 : Functions

Example Question #861 : Differential Functions

Example Question #862 : Functions

Example Question #861 : Differential Functions

Example Question #679 : Other Differential Functions

Example Question #680 : Other Differential Functions

Example Question #861 : Functions

Example Question #682 : How To Find Differential Functions

Example Question #683 : How To Find Differential Functions

All Calculus 1 Resources

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