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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #235 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Calculate the second derivative of the function}\\&-3\cdot 3^{(9x)}\end{align*}$

Possible Answers:

$6561\cdot 3^{(18x)}ln(3)^{3}$

$-27\cdot 3^{(9x)}ln(3)^{2}$

$729\cdot 3^{(18x)}ln(3)^{2}$

$-243\cdot 3^{(9x)}ln(3)^{2}$

Correct answer:

$-243\cdot 3^{(9x)}ln(3)^{2}$

Explanation:

$\begin{align*}&\text{To find the second derivative of a function, one need only be}\\&\text{cautious in performing the steps of derivation twice. For the}\\&\text{function we're given,}\text{ we'll wish to use the following:}\\&d[b^{ax}]=ab^{ax}ln(b)sx\\&\text{Our first derivative is thus:}\\&-27\cdot 3^{(9x)}ln(3)\\&\text{And our second is:}\\&-243\cdot 3^{(9x)}ln(3)^{2}\end{align*}$

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Example Question #435 : Derivative Review

$\begin{align*}&\text{Calculate the second derivative of the function}\\&-19x^{12}\end{align*}$

Possible Answers:

$571824x^{21}$

$-209x^{10}$

$-2508x^{10}$

$51984x^{22}$

Correct answer:

$-2508x^{10}$

Explanation:

$\begin{align*}&\text{To find the second derivative of a function, one need only be}\\&\text{cautious in performing the steps of derivation twice. For the}\\&\text{function we're given,}\text{ we'll wish to use the following:}\\&d[x^a]=ax^{a-1}dx\\&\text{Our first derivative is thus:}\\&-228x^{11}\\&\text{And our second is:}\\&-2508x^{10}\end{align*}$

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Example Question #236 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Take the derivative twice with respect to }x\text{ of the function}\\&y=17\cdot 4^{(2x)}\end{align*}$

Possible Answers:

$68\cdot 4^{(2x)}ln(4)^{2}$

$1156\cdot 4^{(4x)}ln(4)^{2}$

$34\cdot 4^{(2x)}ln(4)^{2}$

$2312\cdot 4^{(4x)}ln(4)^{3}$

Correct answer:

$68\cdot 4^{(2x)}ln(4)^{2}$

Explanation:

$\begin{align*}&\text{To find the second derivative of a function, one need only be}\\&\text{cautious in performing the steps of derivation twice. For the}\\&\text{function we're given,}\text{ we'll wish to use the following:}\\&d[b^{ax}]=ab^{ax}ln(b)sx\\&\text{Our first derivative is thus:}\\&34\cdot 4^{(2x)}ln(4)\\&\text{And our second is:}\\&68\cdot 4^{(2x)}ln(4)^{2}\end{align*}$

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Example Question #1564 : Calculus Ii

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=130ln(9x)e^{(-12x)}\end{align*}$

Possible Answers:

$\frac{(130e^{(-12x)})}{x}- 1560ln(9x)e^{(-12x)}$

$\frac{1}{x}- 12e^{(-12x)}$

$-\frac{(12e^{(-12x)})}{x}$

$\frac{(120e^{(-12x)})}{x}+ 1404ln(9x)e^{(-12x)}$

Correct answer:

$\frac{(130e^{(-12x)})}{x}- 1560ln(9x)e^{(-12x)}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions multiplied together,}\\&\text{the first being: }e^{(-12x)}\\&\text{and the second being: }ln(9x)\\&\text{With the product scaled by }130\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[e^{ax}]=ae^{ax}dx\\&d[ln(ax)]=\frac{dx}{x}\\&u=e^{(-12x)};du=-12e^{(-12x)}\\&v=ln(9x);dv=\frac{1}{x}\\&\text{So the derivative is:}\\&\frac{(130e^{(-12x)})}{x}- 1560ln(9x)e^{(-12x)}\end{align*}$

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Example Question #1565 : Calculus Ii

$\begin{align*}&\text{Calculate the derivative of the function}\\&126x^{17}e^{(-4x)}\end{align*}$

Possible Answers:

$2142x^{16}e^{(-4x)} - 504x^{17}e^{(-4x)}$

$17x^{16} - 4e^{(-4x)}$

$476x^{16}e^{(-4x)} + 1224x^{17}e^{(-4x)}$

$-68x^{16}e^{(-4x)}$

Correct answer:

$2142x^{16}e^{(-4x)} - 504x^{17}e^{(-4x)}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }x^{17}\\&\text{and the second being: }e^{(-4x)}\\&\text{With the product scaled by }126\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[x^a]=ax^{a-1}dx\\&d[e^{ax}]=ae^{ax}dx\\&\text{From the above we can continue:}\\&u=x^{17};du=17x^{16}\\&v=e^{(-4x)};dv=-4e^{(-4x)}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&2142x^{16}e^{(-4x)} - 504x^{17}e^{(-4x)}\end{align*}$

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Example Question #1566 : Calculus Ii

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-70sin(15x)tan(5x)\end{align*}$

Possible Answers:

$- 70sin(15x)\cdot (5tan(5x)^{2} + 5) - 1050cos(15x)tan(5x)$

$-15cos(15x)\cdot (5tan(5x)^{2} + 5)$

$525cos(15x)tan(5x) - 150sin(15x)\cdot (5tan(5x)^{2} + 5)$

$5tan(5x)^{2} - 15cos(15x) + 5$

Correct answer:

$- 70sin(15x)\cdot (5tan(5x)^{2} + 5) - 1050cos(15x)tan(5x)$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }-sin(15x)\\&\text{and the second being: }tan(5x)\\&\text{With the product scaled by }70\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[sin(ax)]=acos(ax)dx\\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\\&\text{From the above we can continue:}\\&u=-sin(15x);du=-15cos(15x)\\&v=tan(5x);dv=5tan(5x)^{2} + 5\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&- 70sin(15x)\cdot (5tan(5x)^{2} + 5) - 1050cos(15x)tan(5x)\end{align*}$

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Example Question #441 : Derivatives

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=170ln(-13x)tan(5x)\end{align*}$

Possible Answers:

$5tan(5x)^{2} +\frac{ 1}{x}+ 5$

$130ln(-13x)\cdot (5tan(5x)^{2} + 5) -\frac{ (85tan(5x))}{x}$

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

$170ln(-13x)\cdot (5tan(5x)^{2} + 5) +\frac{ (170tan(5x))}{x}$

Correct answer:

$170ln(-13x)\cdot (5tan(5x)^{2} + 5) +\frac{ (170tan(5x))}{x}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }ln(-13x)\\&\text{and the second being: }tan(5x)\\&\text{With the product scaled by }170\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\\&\text{From the above we can continue:}\\&u=ln(-13x);du=\frac{1}{x}\\&v=tan(5x);dv=5tan(5x)^{2} + 5\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&170ln(-13x)\cdot (5tan(5x)^{2} + 5) +\frac{ (170tan(5x))}{x}\end{align*}$

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Example Question #441 : Derivative Review

$\begin{align*}&\text{Calculate the derivative of the function}\\&\frac{(100ln(14x))}{x^{20}}\end{align*}$

Possible Answers:

$\frac{100}{x^{21}}-\frac{ (2000ln(14x))}{x^{21}}$

$\frac{1}{x}-\frac{ 20}{x^{21}}$

$-\frac{20}{x^{22}}$

$-\frac{ (2800ln(14x))}{x^{21}}-\frac{ 200}{x^{21}}$

Correct answer:

$\frac{100}{x^{21}}-\frac{ (2000ln(14x))}{x^{21}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }\frac{1}{x^{20}}\\&\text{and the second being: }ln(14x)\\&\text{With the product scaled by }100\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[x^a]=ax^{a-1}dx\\&d[ln(ax)]=\frac{dx}{x}\\&\text{From the above we can continue:}\\&u=\frac{1}{x^{20}};du=-\frac{20}{x^{21}}\\&v=ln(14x);dv=\frac{1}{x}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{100}{x^{21}}-\frac{ (2000ln(14x))}{x^{21}}\end{align*}$

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Example Question #1562 : Calculus Ii

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=28ln(-17x)e^{(-20x)}\end{align*}$

Possible Answers:

$\frac{(28e^{(-20x)})}{x}- 560ln(-17x)e^{(-20x)}$

$-\frac{(20e^{(-20x)})}{x}$

$\frac{1}{x}- 20e^{(-20x)}$

$\frac{(280e^{(-20x)})}{x}- 680ln(-17x)e^{(-20x)}$

Correct answer:

$\frac{(28e^{(-20x)})}{x}- 560ln(-17x)e^{(-20x)}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }ln(-17x)\\&\text{and the second being: }e^{(-20x)}\\&\text{With the product scaled by }28\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[e^{ax}]=ae^{ax}dx\\&\text{From the above we can continue:}\\&u=ln(-17x);du=\frac{1}{x}\\&v=e^{(-20x)};dv=-20e^{(-20x)}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(28e^{(-20x)})}{x}- 560ln(-17x)e^{(-20x)}\end{align*}$

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Example Question #441 : Derivatives

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=72ln(3x)e^{(17x)}\end{align*}$

Possible Answers:

$\frac{(72e^{(17x)})}{x}+ 1224ln(3x)e^{(17x)}$

$17e^{(17x)} +\frac{ 1}{x}$

$\frac{(17e^{(17x)})}{x}$

$-\frac{ (136e^{(17x)})}{x}- 459ln(3x)e^{(17x)}$

Correct answer:

$\frac{(72e^{(17x)})}{x}+ 1224ln(3x)e^{(17x)}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }ln(3x)\\&\text{and the second being: }e^{(17x)}\\&\text{With the product scaled by }72\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[e^{ax}]=ae^{ax}dx\\&\text{From the above we can continue:}\\&u=ln(3x);du=\frac{1}{x}\\&v=e^{(17x)};dv=17e^{(17x)}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(72e^{(17x)})}{x}+ 1224ln(3x)e^{(17x)}\end{align*}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #235 : First And Second Derivatives Of Functions

Example Question #435 : Derivative Review

Example Question #236 : First And Second Derivatives Of Functions

Example Question #1564 : Calculus Ii

Example Question #1565 : Calculus Ii

Example Question #1566 : Calculus Ii

Example Question #441 : Derivatives

Example Question #441 : Derivative Review

Example Question #1562 : Calculus Ii

Example Question #441 : Derivatives

All Calculus 2 Resources

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