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Calculus 1 : Other Differential Functions

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Example Questions

Example Question #61 : Other Differential Functions

Differentiate the function:

$f(x)=\frac{sin(x)}{x}$

Possible Answers:

$f'(x)=\frac{x*cos(x) - sin(x)}{x^2}$

f'(x)=1

$f'(x)=\frac{ sin(x) + cos(x)}{x}$

$f'(x)=\frac{ sin(x) - x*cos(x)}{x^2}$

$f'(x)=\frac{x*sin(x) - cos(x)}{x}$

Correct answer:

$f'(x)=\frac{x*cos(x) - sin(x)}{x^2}$

Explanation:

$f(x)=\frac{sin(x)}{x}$

We evaluate this derivative using the quotient rule:

$if \ f(x)=\frac{g(x)}{h(x)}$ ,

$then\ f'(x)=\frac{g'(x)*h(x) - g(x)*h'(x)}{h(x)^{2}}$ .

Apply the above formula:

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

$f'(x)=\frac{x*cos(x) - sin(x)}{x^2}$ , which is our final answer.

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Example Question #245 : Functions

What is the slope of the line tangent to f(x) = x⁴ – 3x^–4 – 45 at x = 5?

Possible Answers:

355.096

400.096

355.00384

500.00384

422.125

Correct answer:

500.00384

Explanation:

First we must find the first derivative of f(x).

f'(x) = 4x³ + 12x^–5

To find the slope of the tangent line of f(x) at 5, we merely have to evaluate f'(x) at 5:

f'(5) = 4*5³ + 12* 5^–5 = 500 + 12/3125 = 500.00384

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Example Question #63 : Other Differential Functions

Solve for g{}'(x) when

$g(x)=tan^{-1}(4x^3+7x^2)$

Possible Answers:

$g{}'(x)=\frac{12x^2+14x}{1+(4x^3+7x^2)^2}$

$g{}'(x)=sec^2(4x^3+7x^2)$

$g{}'(x)=\frac{4x^3+7x^2}{(12x^2+14x)^2}$

$g{}'(x)=sec^2(12x^2+14x)$

Correct answer:

$g{}'(x)=\frac{12x^2+14x}{1+(4x^3+7x^2)^2}$

Explanation:

$g{}'(x)=tan^{-1}(4x^3+7x^2)$

using the identity:

$\frac{\mathrm{d} }{\mathrm{d} x}\ tan^{-1}[(f(x)]=\frac{f{}'(x)}{1+[f(x)]^2}$

$g{}'(x)=\frac{12x^2+14x}{1+(4x^3+7x^2)^2}$

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Example Question #251 : Differential Functions

Differentiate $\small f(x)=\frac{x^2}{2x+7}$

Possible Answers:

$\small f'(x)=\frac{-2x(x+7)}{(2x+7)^2}$

$\small f'(x)=\frac{2x(x+7)}{(2x+7)}$

$\small f'(x)=\frac{2x(x+7)}{(2x+7)^2}$

$\small f'(x)=\frac{2x(3x+7)}{(2x+7)^2}$

Correct answer:

$\small f'(x)=\frac{2x(x+7)}{(2x+7)^2}$

Explanation:

The Quotient Rule applies when differentiating quotients of functions. Here, $\small f(x)$ equals the quotient of two functions, $\small x^2$ and $\small 2x+7$ . Let $\small l(x)=2x+7$ and $\small h(x)=x^2$ . (Think: $\small l(x)$ is the "low" function or denominator and $\small h(x)$ is the "high" function or numerator.) The Quotient Rule tells us to multiply the "low" function by the derivative of the "high" function, subtract the product of the "high" function and the derivative of the "low" function, and then divide the result by the square of the "low" function. In other words,

$\small \small f'(x)=\frac{l(x)h'(x)-h(x)l'(x)}{[l(x)]^2}$

Here, $\small l(x)=2x+7$ so $\small l'(x)=2$ . Similarly, $\small h(x)=x^2$ so $\small h'(x)=2x$ .

Then

$\small f'(x)=\frac{(2x+7)(2x)-x^2(2)}{(2x+7)^2}$

$\small f'(x)=\frac{(4x^2+14x)-2x^2}{(2x+7)^2}$

$\small f'(x)=\frac{2x^2+14x}{(2x+7)^2}$

Factoring out $\small 2x$ from the numerator gives

$\small \small f'(x)=\frac{2x(x+7)}{(2x+7)^2}$

$\small f'(x)=\frac{-2x(x+7)}{(2x+7)^2}$ inverts the order of the numerator, subtracting $\small l(x)h'(x)$ from $\small h(x)l'(x)$ .

$\small f'(x)=\frac{2x(3x+7)}{(2x+7)^2}$ adds the products in the numerator, rather than subtracting them.

$\small f'(x)=\frac{2x(x+7)}{(2x+7)}$ fails to square the denominator.

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Example Question #64 : How To Find Differential Functions

Differentiate $\small f(x)=\frac{sec(x)}{x}$

Possible Answers:

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

$\small f'(x)=\frac{sec(x)(xtan(x)-1)}{x}$

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

$\small f'(x)=\frac{sec(x)(xtan(x)+1)}{x^2}$

Correct answer:

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

Explanation:

The Quotient Rule applies when differentiating quotients of functions. Here, $\small f(x)$ equals the quotient of two functions, $\small sec(x)$ and $\small x$ . Let $\small l(x)=x$ and $\small h(x)=sec(x)$ . (Think: $\small l(x)$ is the "low" function or denominator and $\small h(x)$ is the "high" function or numerator.) The Quotient Rule tells us to multiply the "low" function by the derivative of the "high" function, subtract the product of the "high" function and the derivative of the "low" function, and then divide the result by the square of the "low" function. In other words,

$\small \small f'(x)=\frac{l(x)h'(x)-h(x)l'(x)}{[l(x)]^2}$

Here, $\small l(x)=x$ so $\small l'(x)=1$ . Similarly, $\small h(x)=sec(x)$ so $\small h'(x)=sec(x)tan(x)$ .

Then

$\small f'(x)=\frac{(x)[sec(x)tan(x)]-[sec(x)](1)}{x^2}$

$\small f'(x)=\frac{xsec(x)tan(x)-sec(x)}{x^2}$

Factoring out $\small sec(x)$ gives

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

$\small f'(x)=\frac{sec(x)(1-xtan(x))}{x^2}$ inverts the order of the numerator, subtracting $\small l(x)h'(x)$ from $\small h(x)l'(x)$ .

$\small f'(x)=\frac{sec(x)(xtan(x)+1)}{x^2}$ adds the products in the numerator, rather than subtracting them.

$\small f'(x)=\frac{sec(x)(xtan(x)-1)}{x}$ fails to square the denominator.

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Example Question #251 : Functions

Differentiate $\small f(x)=\frac{x-1}{x+5}$

Possible Answers:

$\small \small \small f'(x)=\frac{-6}{(x+5)^2}$

$\small \small f'(x)=\frac{6}{(x+5)^2}$

$\small \small f'(x)=\frac{6}{(x+5)}$

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

Correct answer:

$\small \small f'(x)=\frac{6}{(x+5)^2}$

Explanation:

The Quotient Rule applies when differentiating quotients of functions. Here, $\small f(x)$ equals the quotient of two functions, $\small x-1$ and $\small x+5$ . Let $\small \small \small \small \small l(x)=x+5$ and $\small \small h(x)=x-1$ . (Think: $\small l(x)$ is the "low" function or denominator and $\small h(x)$ is the "high" function or numerator.) The Quotient Rule tells us to multiply the "low" function by the derivative of the "high" function, subtract the product of the "high" function and the derivative of the "low" function, and then divide the result by the square of the "low" function. In other words,

$\small \small f'(x)=\frac{l(x)h'(x)-h(x)l'(x)}{[l(x)]^2}$

Here, $\small l(x)=x+5$ so $\small l'(x)=1$ . Similarly, $\small h(x)=x-1$ so $\small h'(x)=1$ .

Then

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

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

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

$\small \small f'(x)=\frac{6}{(x+5)^2}$

$\small \small \small f'(x)=\frac{-6}{(x+5)^2}$ inverts the order of the numerator, subtracting $\small l(x)h'(x)$ from $\small h(x)l'(x)$ .

$\small \small \small f'(x)=\frac{2x+4}{(x+5)^2}$ adds the products in the numerator, rather than subtracting them.

$\small \small f'(x)=\frac{6}{(x+5)}$ fails to square the denominator.

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Example Question #66 : How To Find Differential Functions

Differentiate $\small f(x)=\sqrt{x^2+1}$

Possible Answers:

$\small f'(x)=x^3+x$

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

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

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

Correct answer:

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

Explanation:

The Chain Rule applies when differentiating compositions of functions. Here, $\small \small f(x)$ equals the composition of two functions, $\small \sqrt{x}$ and $\small x^2+1$ . Let $\small g(x)=\sqrt{x}$ and $\small h(x)=x^2+1$ . Then $\small \small f(x)=g(h(x))$ and the Chain Rule tells us to differentiate the outside function $\small \small g(x)$ and multiply the result by the derivative of the inside function $\small \small h(x)$ . In other words, $\small \small f'(x)=g'(h(x))h'(x)$ . Note that the inside function $\small \small h(x)$ is left untouched when the outside function $\small \small g(x)$ is differentiated. Here, $\small g'(h(x))=\frac{1}{2}(x^2+1)^{-\frac{1}{2}}$ and $\small h'(x)=2x$ . Remember, roots can (and should) be rewritten as fractional exponents, so $\small \sqrt{x^2+1}$ becomes $\small (x^2+1)^{\frac{1}{2}}$ which is then differentiated like any other exponent. So

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

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

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

$\small f'(x)=\frac{1}{2\sqrt{x^2+1}}$ is an incomplete application of the Chain Rule which neglects to multiply the derivative of the outside function by the derivative of the inside function.

$\small f'(x)=\frac{1}{2\sqrt{x^2+1}}+2x$ is a misapplication of the Chain Rule which adds the derivative of the outside and inside functions rather than multiplying them.

$\small f'(x)=x^3+x$ is a misapplication of the Power Rule which fails to subtract 1 from the original exponent.

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Example Question #67 : How To Find Differential Functions

Differentiate f(x)=cos^2(x)

Possible Answers:

f'(x)=-2sin(x)

f'(x)=2cos(x)-sin(x)

f'(x)=-2sin(x)cos(x)

f'(x)=2cos(x)

Correct answer:

f'(x)=-2sin(x)cos(x)

Explanation:

The Chain Rule applies when differentiating compositions of functions. Here, $\small \small f(x)$ equals the composition of two functions, $\small \small x^2$ and $\small \small cos(x)$ . Let $\small \small \small g(x)=x^2$ and $\small \small \small h(x)=cos(x)$ . Then $\small \small f(x)=g(h(x))$ and the Chain Rule tells us to differentiate the outside function $\small \small g(x)$ and multiply the result by the derivative of the inside function $\small \small h(x)$ . In other words, $\small \small f'(x)=g'(h(x))h'(x)$ . Note that the inside function $\small \small h(x)$ is left untouched when the outside function $\small \small g(x)$ is differentiated. Here, $\small \small g'(h(x))=2cos(x)$ and $\small \small h'(x)=-sin(x)$ , so $\small f'(x)=(2cos(x))(-sin(x))$ which simplifies to $\small \small \small f'(x)=-2sin(x)cos(x)$ .

$\small f'(x)=2cos(x)$ is an incomplete application of the Chain Rule which neglects to multiply the derivative of the outside function by the derivative of the inside function.

$\small f'(x)=2cos(x)-sin(x)$ is a misapplication of the Chain Rule which adds the derivative of the outside and inside functions rather than multiplying them.

$\small f'(x)=-2sin(x)$ is a misapplication of the Chain Rule which substitutes the derivative of the inside function for the original inside function rather than multiplying the derivative of the outside function by the derivative of the inside function.

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Example Question #68 : How To Find Differential Functions

Differentiate $\small f(x)=sin(6x)$

Possible Answers:

$\small f'(x)=cos(6x)+6$

$\small f'(x)=6cos(6x)$

$\small f'(x)=6cos(x)$

$\small f'(x)=cos(6x)$

Correct answer:

$\small f'(x)=6cos(6x)$

Explanation:

The Chain Rule applies when differentiating compositions of functions. Here, $\small \small f(x)$ equals the composition of two functions, $\small sin(x)$ and $\small 6x$ . Let $\small \small \small \small g(x)=sin(x)$ and $\small \small \small \small h(x)=6x$ . Then $\small \small f(x)=g(h(x))$ and the Chain Rule tells us to differentiate the outside function $\small \small g(x)$ and multiply the result by the derivative of the inside function $\small \small h(x)$ . In other words, $\small \small f'(x)=g'(h(x))h'(x)$ . Note that the inside function $\small \small h(x)$ is left untouched when the outside function $\small \small g(x)$ is differentiated. Here, $\small \small \small g'(h(x))=cos(6x)$ and $\small \small \small h'(x)=6$ , so $\small \small f'(x)=(cos(6x))(6)$ which simplifies to $\small \small \small \small f'(x)=6cos(6x)$ .

$\small \small f'(x)=cos(6x)$ is an incomplete application of the Chain Rule which neglects to multiply the derivative of the outside function by the derivative of the inside function.

$\small \small f'(x)=cos(6x)+6$ is a misapplication of the Chain Rule which adds the derivative of the outside and inside functions rather than multiplying them.

$\small \small f'(x)=6cos(x)$ is a misapplication of the Chain Rule which fails to preserve the original inside function when differentiating the outside function.

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Example Question #69 : How To Find Differential Functions

Differentiate $\small f(x)=(3x+2)^5$

Possible Answers:

$\small f'(x)=5(3x+2)^4$

$\small f'(x)=5(3x+2)^4+3x$

$\small f'(x)=5(3x+2)^4+3$

$\small f'(x)=15(3x+2)^4$

Correct answer:

$\small f'(x)=15(3x+2)^4$

Explanation:

The Chain Rule applies when differentiating compositions of functions. Here, $\small \small f(x)$ equals the composition of two functions, $\small x^5$ and $\small 3x+2$ . Let $\small \small \small \small g(x)=x^5$ and $\small \small \small \small h(x)=3x+2$ . Then $\small \small f(x)=g(h(x))$ and the Chain Rule tells us to differentiate the outside function $\small \small g(x)$ and multiply the result by the derivative of the inside function $\small \small h(x)$ . In other words, $\small \small f'(x)=g'(h(x))h'(x)$ . Note that the inside function $\small \small h(x)$ is left untouched when the outside function $\small \small g(x)$ is differentiated. Here, $\small \small \small \small g'(h(x))=5(3x+2)^4$ and $\small \small \small h'(x)=3$ , so $\small \small \small \small f'(x)=(5(3x+2)^4)(3)$ which simplifies to $\small \small \small \small f'(x)=15(3x+2)^4$ .

$\small f'(x)=5(3x+2)^4$ is an incomplete application of the Chain Rule which neglects to multiply the derivative of the outside function by the derivative of the inside function.

$\small f'(x)=5(3x+2)^4+3$ is a misapplication of the Chain Rule which adds the derivative of the outside and inside functions rather than multiplying them.

$\small f'(x)=5(3x+2)^4+3x$ is a misapplication of the Chain Rule which adds the derivative of the outside function to an incorrect derivation of the inside function.

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Calculus 1 : Other Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #61 : Other Differential Functions

Example Question #245 : Functions

Example Question #63 : Other Differential Functions

Example Question #251 : Differential Functions

Example Question #64 : How To Find Differential Functions

Example Question #251 : Functions

Example Question #66 : How To Find Differential Functions

Example Question #67 : How To Find Differential Functions

Example Question #68 : How To Find Differential Functions

Example Question #69 : How To Find Differential Functions

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