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Calculus 1 : How to find differential functions

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1646 : Calculus

Find the derivative.

x^4+2x-3x^3

Possible Answers:

-4x^3-9x^2+2

4x^3-9x^2-2

4x^3+9x^2+2

4x^3-9x^2+2

Correct answer:

4x^3-9x^2+2

Explanation:

Use the power rule to find the derivative.

Remember the power rule:

$(x^n)'=nx^{n-1}$

We can now apply this to our situation.

$\frac{d}{dx}x^4=4x^3$

$\frac{d}{dx}-3x^3=-9x^2$

$\frac{d}{dx}2x=2$

The derivative is 4x^3-9x^2+2

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Example Question #1647 : Calculus

Find the derivative at x=4 .

$x^{2}+8x^3-2x$

Possible Answers:

394

270

400

308

Correct answer:

394

Explanation:

First, use the power rule to find the derivative.

Remember the power rule:

$(x^n)'=nx^{n-1}$

We can now apply this to our situation.

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

$\frac{d}{dx}8x^3=24x^2$

$\frac{d}{dx}2x=2$

The derivative is 24x^2+2x+2

Now, substitute for .

24(4^2)+2(4)+2

=394

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Example Question #1648 : Calculus

Find the derivative at x=5 .

x^3-2x^2+9x

Possible Answers:

Correct answer:

Explanation:

First, find the derivative using the power rule.

Remember the power rule:

$(x^n)'=nx^{n-1}$

We can now apply this to our situation.

$\frac{d}{dx}x^3=3x^2$

$\frac{d}{dx}-2x^2=-4x$

$\frac{d}{dx}9x=9$

The derivative is 3x^2-4x+9

Now, substitute for .

3(5^2)-4(5)+9

=64

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Example Question #431 : How To Find Differential Functions

Find the derivative.

7x^4+8x^3+9x^2+10x+1

Possible Answers:

28x^3+24x^2+18x+10

28x^3+24x^2+28x

28x^3+24x^2+18x

28x^3

Correct answer:

28x^3+24x^2+18x+10

Explanation:

Find the derivative using the power rule.

Remember the power rule:

$(x^n)'=nx^{n-1}$

We can now apply this to our situation.

$\frac{d}{dx}7x^4=28x^3$

$\frac{d}{dx}8x^3=24x^2$

$\frac{d}{dx}9x^2=18x$

$\frac{d}{dx}10x=10$

$\frac{d}{dx}1=0$

The derivative is 28x^3+24x^2+18x+10

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Example Question #1651 : Calculus

Find the derivative at x=6 .

7x^2-4x

Possible Answers:

Correct answer:

Explanation:

First, find the derivative using the power rule.

Remember the power rule:

$(x^n)'=nx^{n-1}$

We can now apply this to our situation.

$\frac{d}{dx}7x^2=14x$

$\frac{d}{dx}-4x=-4$

The derivative is 14x-4

Now, substitute for .

14(6)-4=80

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Example Question #1652 : Calculus

Find the derivative.

$5x\cos (x)$

Possible Answers:

$5x\sin (x)+5\cos (x)$

$-5x\sin (x)-5\cos (x)$

$-5x\sin (x)+5\cos (x)$

$-x\sin (x)+5\cos (x)$

Correct answer:

$-5x\sin (x)+5\cos (x)$

Explanation:

Use the product rule to find the derivative.

Remember the product rule:

(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)

We can now apply this to our situation.

$5x(-\sin(x))+\cos (x)(5)$

Simplify.

$-5x\sin (x)+5\cos (x)$

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Example Question #1653 : Calculus

Find the derivative.

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

Possible Answers:

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

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

Correct answer:

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

Explanation:

Use the quotient rule to find the derivative.

Remember the quotient rule:

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

We can now apply this to our situation.

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

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

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Example Question #432 : How To Find Differential Functions

Find the derivative.

$\frac{5}{x^2}$

Possible Answers:

$\frac{5}{x^4}$

$\frac{10}{x^4}$

$\frac{10}{x^3}$

$\frac{5}{x^3}$

Correct answer:

$\frac{10}{x^3}$

Explanation:

Find the derivative using the quotient rule.

Remember the quotient rule:

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

We can now apply this to our situation.

$\frac{x^2(0)-5(2x)}{x^4}$

$=\frac{10x}{x^4}=\frac{10}{x^3}$

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Example Question #1655 : Calculus

Find the derivative.

$\frac{x^2}{3}$

Possible Answers:

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

$\frac{x}{9}$

$\frac{2x}{3}$

$\frac{2x}{9}$

Correct answer:

$\frac{2x}{3}$

Explanation:

Find the derivative using the quotient rule.

Remember the quotient rule:

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

We can now apply this to our situation.

$\frac{d}{dx}\frac{x^2}{3}=\frac{3(2x)-x^2(0)}{9}$

$=\frac{6x}{9}=\frac{2x}{3}$

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Example Question #1652 : Calculus

Let f(x)=x^5+7x on the interval (0,2) . Find a value for the number(s) that satisfies the mean value theorem for this function and interval.

Possible Answers:

-0.89

-1.17

1.17

1.33

0.89

Correct answer:

1.33

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^5+7x on the interval (0,2)

f(0)=0^5+7(0)=0

f(2)=2^5+7(2)=46

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

$\frac{46-0}{2-0}=23$

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

f'(x)=5x^4+7

5x^4+7=23

x=-1.33,1.33

The solution x=1.33 validates the mean value theorem by falling within (0,2)

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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 #1646 : Calculus

Example Question #1647 : Calculus

Example Question #1648 : Calculus

Example Question #431 : How To Find Differential Functions

Example Question #1651 : Calculus

Example Question #1652 : Calculus

Example Question #1653 : Calculus

Example Question #432 : How To Find Differential Functions

Example Question #1655 : Calculus

Example Question #1652 : Calculus

All Calculus 1 Resources

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