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Precalculus : Pre-Calculus

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12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #61 : Derivatives

Find the first derivative of $y=x^{3}$ .

Possible Answers:

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

$\frac{dy}{dx}=3x$

$y=3x^{2}$

$\frac{dx}{dy}=3x$

y=3x

Correct answer:

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

Explanation:

By the Power Rule of derivatives, for any equation $y=x^{a}$ , the derivative $\frac{dy}{dx}=a\cdot x^{a-1}$ .

With our function $y=x^{3}$ , where a=3 , we can therefore conclude that:

$\frac{dy}{dx}=3\cdot x^{3-1}$

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

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Example Question #62 : Derivatives

Find the derivative of $y=4x^{5}$ .

Possible Answers:

$\frac{dy}{dx}=20x$

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

$\frac{dy}{dx}=20x^{4}$

$\frac{dy}{dx}=9x^{4}$

$y=20x^{4}$

Correct answer:

$\frac{dy}{dx}=20x^{4}$

Explanation:

By the Power Rule of derivatives, for any equation $y=x^{a}$ , the derivative $\frac{dy}{dx}=a\cdot x^{a-1}$ .

Given our function $y=4x^{5}$ , where a=5 , we can conclude that

$\frac{dy}{dx}=5(4\cdot x^{5-1})$

$\frac{dy}{dx}=20x^{4}$

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Example Question #63 : Derivatives

Find the first derivative of the following function:

$f(x)=5x^{3}+3x^{2}-\frac{1}{2}x+9$

Possible Answers:

$f'(x)=15x^{2}+6x-\frac{1}{2}$

$f'(x)=8x^{2}+5x-\frac{3}{2}$

$f(x)=15x^{2}+5x-\frac{1}{2}$

$f'(x)=15x^{2}+5x-\frac{17}{2}$

$f'(x)=21x-\frac{1}{2}$

Correct answer:

$f'(x)=15x^{2}+6x-\frac{1}{2}$

Explanation:

To take the derivative of a polynomial, we'll need to apply the power rule to a term with a coeffient and an exponent :

$\frac{d}{dx}(Ax^{n})=nAx^{n-1}$

Applying this rule to each term in the polynomial:

$f(x)=5x^{3}+3x^{2}-\frac{1}{2}x+9$

$f'(x)=3(5x^{3-1})+2(3x^{2-1})-1\left(\frac{1}{2}x^{1-1}\right)+0(9)$

$f'(x)=15x^{2}+6x-\frac{1}{2}$

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Example Question #732 : Pre Calculus

Find the derivative of the function $f(x)=x^{3}+5$ .

Possible Answers:

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

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

$f'(x)=\frac{1}4{}x^{4}$

None of the above

$f'(x)=3x^{2}+5$

Correct answer:

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

Explanation:

For any function $f(x)=Ax^{n}$ , the first derivative $f'(x)=nAx^{n-1}$ .

Therefore, taking each term of f(x) :

$f(x)=x^{3}+5$

$f'(x)=3x^{3-1}+0(5)$

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

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Example Question #21 : Find The First Derivative Of A Function

Find the derivative of $f(x)=\frac{x^{2}}{2}-x$ .

Possible Answers:

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

f'(x)=x-1

f'(x)=x+1

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

f'(x)=1-x

Correct answer:

f'(x)=x-1

Explanation:

For any function $f(x)=Ax^{n}$ , the first derivative $f'(x)=nAx^{n-1}$ .

Therefore, taking each term of f(x) :

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

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

f'(x)=x-1

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Example Question #22 : Find The First Derivative Of A Function

Find the derivative of the following function:

$f(x)=3(2x^3+\cos (x))^{50}$

Possible Answers:

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

$f'(x)=150(2x^3+\cos (x))^{49}$

$f'(x)=150(2x^3+\cos (x))^{49}(6x^2-\sin(x))$

$f'(x)=150(2x^3+\cos (x))^{49}(6x^2-\sin(x))(12x-\cos(x))$

$f'(x)=150(6x^2-\sin(x))^{49}$

Correct answer:

$f'(x)=150(2x^3+\cos (x))^{49}(6x^2-\sin(x))$

Explanation:

First apply the power rule to the entire function, multiplying it by its exponent and then subtracting 1 from the new exponent. Then apply the chain rule, multiplying the expression obtained from the power rule by the derivative of just the term inside of the parentheses. Remember the derivative of cos(x) is -sin(x):

$f(x)=3(2x^3+\cos (x))^{50}$

$f'(x)=150(2x^3+\cos (x))^{49}\frac{d}{dx}(2x^3+\cos(x))$

$f'(x)=150(2x^3+\cos (x))^{49}(6x^2-\sin(x))$

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Example Question #23 : Find The First Derivative Of A Function

Find the derivative of the following function

f(x)=-3x^4+6x^3-x^2+8x-17

Possible Answers:

f'(x)=-9x^3+12x^2-x

$f'(x)=-\frac{3}{4}x^3+2x^2-\frac{1}{2}x+8$

$f'(x)=-12x^4+18x^3-2x^2+8x\textup{}$

f'(x)=-12x^3+18x^2-2x+8

f'(x)=-3x^3+6x^2-x+8

Correct answer:

f'(x)=-12x^3+18x^2-2x+8

Explanation:

To find the derivative of this function, we simply need to use the Power Rule. The Power rule states that for each term, we simply multiply the coefficient by the power to find the new coefficient. We then decrease the power by one to obtain the degree of the new term.

For example, with our first term, -3x^4 , we would multiply the coefficient (-3) by the power (4) to obtain the new coefficient of . We then decrease the power by one from 4 to 3 for the new degree. Therefore, our new term is -12x^3 . We then simply repeat the process with the remaining terms.

Note that with the second to last term, our degree is 1. Therefore, multiplying the coefficient by the power gives us the same coefficient of 8. When the degree decreases by one, we have a degree of 0, which simply becomes 1, making the entire term simply 8.

8x^0=8(1)=8

With our final term, we technically have

17=17x^0

Therefore, multiplying our coefficient by our power of 0 makes the whole term 0 and thus negligible.

Our final derivative then is

f'(x)=-12x^3+18x^2-2x+8

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Example Question #24 : Find The First Derivative Of A Function

Find the first derivative of the following function.

f(x)=3x^2+4

Possible Answers:

f'(x)=6x

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

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

$f'(x)=6x-\frac{1}{x}+xe^x$

Correct answer:

f'(x)=6x

Explanation:

To find the first derivative, we will use the definition of the derivative which is,

$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ .

In our case our,

f(x)=3x^2+4 and

f(x+h)=3(x+h)^2+4=3(x^2+2xh+h^2)+4 .

Plugging these functions into our formula we get the following.

$f'(x)=\lim_{h\rightarrow 0}\frac{3(x^2+2xh+h^2)+4-3x^2-4}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{3x^2+6xh+3h^2+4-3x^2-4}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{6xh+3h^2}{h}$

$f'(x)=\lim_{h\rightarrow 0}\frac{h}{h}\frac{6x+3h}{1}$

$f'(x)=\lim_{h\rightarrow 0}{6x+3h}$

Now plugging in zero for h we are able to find the derivative of our function.

$f'(x)=\lim_{h\rightarrow 0}{6x+3(0)}=6x$

f'(x)=6x

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Example Question #735 : Pre Calculus

Find where y=4x^5+3x^3+6x^2+1 .

Possible Answers:

y'=20x^4+9x^2+12x

y'=20x^4+9x^2+12x+1

y'=20x^5+9x^3+12x^2

y'=9x^4+6x^2+8x+1

Correct answer:

y'=20x^4+9x^2+12x

Explanation:

In order to find the derivative we will need to use the power rule on each term. The power rule states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Applying this rule we get the following.

y=4x^5+3x^3+6x^2+1

$y'=(4*5)x^{5-1}+(3*3)x^{3-1}+(6*2)x^{2-1}+(1)'$

y'=20x^4+9x^2+12x+0

y'=20x^4+9x^2+12x

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Example Question #25 : Find The First Derivative Of A Function

Find f'(x) if $f(x)=\frac{5x^2}{3x+2}$ .

Possible Answers:

5x(x+2)

$\frac{5x(3x+4)}{(3x+2)^2}$

$\frac{5x(3x+4)}{3x+2}$

$\frac{5x(9x+4)}{(3x+2)^2}$

Correct answer:

$\frac{5x(3x+4)}{(3x+2)^2}$

Explanation:

Because the original function is the quotient of 2 separate functions, we can use the Quotient Rule.

Quotient Rule:

$f(x)=\frac{u(x)}{v(x)}$ ,

then

$f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{(v(x))^2}$ .

$f(x)=\frac{5x^2}{3x+2}$ gives us

u(x)=5x^2 , and v(x)=3x+2 .

Now

$f'(x)=\frac{(5x^2)'(3x+2)-(5x^2)(3x+2)'}{(3x+2)^2}$

$f'(x)=\frac{10x(3x+2)-(5x^2)(3)}{(3x+2)^2}$ .

Simplifying yields

$f'(x)=\frac{30x^2+20x-15x^2}{(3x+2)^2}$ or

$f'(x)=\frac{15x^2+20x}{(3x+2)^2}$ .

We can factor out to see that finally

$f'(x)=\frac{5x(3x+4)}{(3x+2)^2}$ .

Because the two terms 3x+4 and 3x+2 are different, no more terms can cancel.

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #61 : Derivatives

Example Question #62 : Derivatives

Example Question #63 : Derivatives

Example Question #732 : Pre Calculus

Example Question #21 : Find The First Derivative Of A Function

Example Question #22 : Find The First Derivative Of A Function

Example Question #23 : Find The First Derivative Of A Function

Example Question #24 : Find The First Derivative Of A Function

Example Question #735 : Pre Calculus

Example Question #25 : Find The First Derivative Of A Function

All Precalculus Resources

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