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Calculus 1 : How to find differential functions

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Example Questions

Example Question #721 : How To Find Differential Functions

What is the derivative of $f(x)=\frac{4}{3}x^3-5x^2$ ?

Possible Answers:

f{}'(x)=4x-10x

$f{}'(x)=4x^2-10x$

$f{}'(x)=4x^2-10$

$f{}'(x)=4x^2-1$

$f{}'(x)=4x^2+10x$

Correct answer:

$f{}'(x)=4x^2-10x$

Explanation:

When taking the derivative, remember to multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent. Therefore, after differentiating each term, you get: $f{}'(x)=4x^2-10x$ .

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Example Question #722 : How To Find Differential Functions

Let $f(x)=xe^{4x}$ 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:

0.67

0.45

0.23

0.18

0.92

Correct answer:

0.67

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 $f(x)=xe^{4x}$ on the interval (0,1)

$f(0)=(0)e^{4(0)}=0$

$f(1)=(1)e^{4(1)}=54.6$

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

$\frac{54.6-0}{1-0}=54.6$

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

Derivative of an exponential:

d[e^u]=e^udu

Derivative of a natural log:

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

Product rule: d[uv]=udv+vdu

$f'(x)=e^{4x}+4xe^{4x}$

$e^{4x}+4xe^{4x}=54.6$

Using a calculator, we find the solution x=0.67 , which fits within the interval (0,1) , satisfying the mean value theorem.

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Example Question #723 : How To Find Differential Functions

Find the derivative.

5x^2-6x+3

Possible Answers:

5x-3

10x-6

5x-6

10x-3

Correct answer:

10x-6

Explanation:

Use the power rule to find the derivative.

$\frac{d}{dx}5x^2=10x$

$\frac{d}{dx}-6x=-6$

$\frac{d}{dx}3=0$

Thus, the derivative is 10x-6

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Example Question #723 : Other Differential Functions

Find the derivative.

x^4-x^3+x

Possible Answers:

4x^3-3x^2+1

x^3+x^2+1

x^4-x^3+x

x^3-x^2

Correct answer:

4x^3-3x^2+1

Explanation:

Use the power rule to find the derivative.

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

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

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

Thus, the derivative is 4x^3-3x^2+1 .

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Example Question #912 : Differential Functions

Find the derivative:

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

Possible Answers:

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

Answer not listed

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

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

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

Correct answer:

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

Explanation:

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

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

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)= ln(2x+e^x)

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

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

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Example Question #724 : How To Find Differential Functions

Find the derivative of the following function:

$f(x)= 4e^{2x} + 3x^4$

Possible Answers:

f'(x)= 8 + 12x

$f(x)= e^{x} + x^4$

$f'(x)= 8e^{2x} + 12x^{3}$

Answer not listed

$f(x)= e^{x} + 3x^3$

Correct answer:

$f'(x)= 8e^{2x} + 12x^{3}$

Explanation:

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

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

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: $f(x)= 4e^{2x} + 3x^4$

That is done by doing the following: $f'(x)= 4e^{2x}(2) + (4)3x^{4-1}$

Therefore, the answer is: $f'(x)= 8e^{2x} + 12x^{3}$

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Example Question #724 : How To Find Differential Functions

Find the first derivative of x^3-15x^2+75x-45 .

Possible Answers:

x^2-15x+75

3x^2+6x-12

3x^2-30x+30

3x^2-30x+75

x^2-15x+30

Correct answer:

3x^2-30x+75

Explanation:

We need to differentiate term by term, applying the power rule,

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

This gives us

$\begin{align*} (x^3-15x^2+75x-45)' &= 3x^2-30x+75 \end{align*}$

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Example Question #725 : How To Find Differential Functions

Find the first derivative of x^3-3x^2-2 .

Possible Answers:

None of the other answers.

x^2-3x-2

3x^2-6x-2

x^2-3x

3x^2-6x

Correct answer:

3x^2-6x

Explanation:

We need to differentiate term by term, applying the power rule,

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

This gives us

$\begin{align*} (x^3-3x^2-2)' &= 3x^2-6x \end{align*}$

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Example Question #727 : How To Find Differential Functions

Find the dervivative: f(x)= (x^4 + ln(x) - 3)^2

Possible Answers:

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

Answer not listed

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

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

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

Correct answer:

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

Explanation:

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

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

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)= (x^4 + ln(x) - 3)^2

That is done by doing the following: $f'(x)= (2)(x^4 + ln(x) - 3)^{(2-1)} (4x^{(4-1)} + \frac{1}{x} - 0)$

Therefore, the answer is: $f'(x)= 2(x^4 + ln(x) - 3) (4x^{3} + \frac{1}{x} )$

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Example Question #728 : How To Find Differential Functions

Find the derivative: $f(x)= \sin(2x^4 - e^{2x})$

Possible Answers:

$f'(x)= -\cos(2x^4 - e^{2x}) (8x^{3} - e^{2x})$

$f'(x)= \sin(2x^4 - e^{x}) (8x^{3} - e^{x})$

$f'(x)= \cos(2x^4 - e^{2x}) (8x^{3} - 2e^{2x})$

Answer not listed

$f'(x)= -\cos(8x^3 - e^{2x}) (8x^{3} - e^{2x})$

Correct answer:

$f'(x)= \cos(2x^4 - e^{2x}) (8x^{3} - 2e^{2x})$

Explanation:

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

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

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: $f(x)= \sin(2x^4 - e^{2x})$

That is done by doing the following: $f'(x)= \cos(2x^4 - e^{2x}) ((4)2x^{(4-1)} - e^{2x} (2))$

Therefore, the answer is: $f'(x)= \cos(2x^4 - e^{2x}) (8x^{3} - 2e^{2x})$

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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 #721 : How To Find Differential Functions

Example Question #722 : How To Find Differential Functions

Example Question #723 : How To Find Differential Functions

Example Question #723 : Other Differential Functions

Example Question #912 : Differential Functions

Example Question #724 : How To Find Differential Functions

Example Question #724 : How To Find Differential Functions

Example Question #725 : How To Find Differential Functions

Example Question #727 : How To Find Differential Functions

Example Question #728 : How To Find Differential Functions

All Calculus 1 Resources

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