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Calculus 2 : First and Second Derivatives of Functions

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Example Questions

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:

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

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

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

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

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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Example Question #241 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Calculate the derivative of the function}\\&-180x^{4}ln(13x)\end{align*}$

Possible Answers:

$80x^{3} - 468x^{3}ln(13x)$

$\frac{1}{x}+ 4x^{3}$

$- 720x^{3}ln(13x) - 180x^{3}$

$4x^{2}$

Correct answer:

$- 720x^{3}ln(13x) - 180x^{3}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }ln(13x)\\&\text{and the second being: }x^{4}\\&\text{With the product scaled by }-180\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[x^a]=ax^{a-1}dx\\&\text{From the above we can continue:}\\&u=ln(13x);du=\frac{1}{x}\\&v=x^{4};dv=4x^{3}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&- 720x^{3}ln(13x) - 180x^{3}\end{align*}$

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Example Question #442 : Derivative Review

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=-42cos(12x)ln(19x)\end{align*}$

Possible Answers:

$504ln(19x)sin(12x) -\frac{ (42cos(12x))}{x}$

$\frac{(3024cos(12x))}{x}- 67032ln(19x)sin(12x)$

$504sin(12x) -\frac{ 42}{x}$

$\frac{(504sin(12x))}{x}$

Correct answer:

$504ln(19x)sin(12x) -\frac{ (42cos(12x))}{x}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }ln(19x)\\&\text{and the second being: }cos(12x)\\&\text{With the product scaled by }-42\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[cos(ax)]=-asin(ax)dx\\&\text{From the above we can continue:}\\&u=ln(19x);du=\frac{1}{x}\\&v=cos(12x);dv=-12sin(12x)\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&504ln(19x)sin(12x) -\frac{ (42cos(12x))}{x}\end{align*}$

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Example Question #241 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Calculate the derivative of the function}\\&-98sin(4x)tan(14x)\end{align*}$

Possible Answers:

$98tan(14x)^{2} - 56cos(4x) + 98$

$392sin(4x)\cdot (14tan(14x)^{2} + 14) + 5488cos(4x)tan(14x)$

$-4cos(4x)\cdot (1372tan(14x)^{2} + 1372)$

$- 98sin(4x)\cdot (14tan(14x)^{2} + 14) - 392cos(4x)tan(14x)$

Correct answer:

$- 98sin(4x)\cdot (14tan(14x)^{2} + 14) - 392cos(4x)tan(14x)$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }-tan(14x)\\&\text{and the second being: }-sin(4x)\\&\text{With the product scaled by }-98\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\\&d[sin(ax)]=acos(ax)dx\\&\text{From the above we can continue:}\\&u=-tan(14x);du=- 14tan(14x)^{2} - 14\\&v=-sin(4x);dv=-4cos(4x)\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&- 98sin(4x)\cdot (14tan(14x)^{2} + 14) - 392cos(4x)tan(14x)\end{align*}$

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Example Question #444 : Derivative Review

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=\frac{(120x^{14})}{4^{(4x)}}\end{align*}$

Possible Answers:

$-\frac{ (6720x^{13})}{4^{(4x)}}-\frac{ (6720x^{14}ln(4))}{4^{(4x)}}$

$-\frac{(6720x^{13}ln(4))}{4^{(4x)}}$

$\frac{(40ln(4))}{4^{(4x)}}- 168x^{13}$

$\frac{(1680x^{13})}{4^{(4x)}}-\frac{ (480x^{14}ln(4))}{4^{(4x)}}$

Correct answer:

$\frac{(1680x^{13})}{4^{(4x)}}-\frac{ (480x^{14}ln(4))}{4^{(4x)}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }x^{14}\\&\text{and the second being: }\frac{1}{4^{(4x)}}\\&\text{With the product scaled by }120\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[x^a]=ax^{a-1}dx\\&d[b^{ax}]=ab^{ax}ln(b)sx\\&\text{From the above we can continue:}\\&u=x^{14};du=14x^{13}\\&v=\frac{1}{4^{(4x)}};dv=-\frac{(4ln(4))}{4^{(4x)}}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&\frac{(1680x^{13})}{4^{(4x)}}-\frac{ (480x^{14}ln(4))}{4^{(4x)}}\end{align*}$

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Example Question #445 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=\frac{(108e^{(-4x)})}{3^{(7x)}}\end{align*}$

Possible Answers:

$-\frac{ (432e^{(-4x)})}{3^{(7x)}}-\frac{ (756e^{(-4x)}ln(3))}{3^{(7x)}}$

$\frac{(3024e^{(-4x)})}{3^{(7x)}}+\frac{ (3024e^{(-4x)}ln(3))}{3^{(7x)}}$

$- 36e^{(-4x)} -\frac{ (84ln(3))}{3^{(7x)}}$

$\frac{(3024e^{(-4x)}ln(3))}{3^{(7x)}}$

Correct answer:

$-\frac{ (432e^{(-4x)})}{3^{(7x)}}-\frac{ (756e^{(-4x)}ln(3))}{3^{(7x)}}$

Explanation:

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions}\\&\text{multiplied together, the first being: }e^{(-4x)}\\&\text{and the second being: }\frac{1}{3^{(7x)}}\\&\text{With the product scaled by }108\\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\\&\text{along with the following:}\\&d[e^{ax}]=ae^{ax}dx\\&d[b^{ax}]=ab^{ax}ln(b)sx\\&\text{From the above we can continue:}\\&u=e^{(-4x)};du=-4e^{(-4x)}\\&v=\frac{1}{3^{(7x)}};dv=-\frac{(7ln(3))}{3^{(7x)}}\\&\text{We can then combine these terms (Don't forget the scalar!)}\\&\text{So the derivative is:}\\&-\frac{ (432e^{(-4x)})}{3^{(7x)}}-\frac{ (756e^{(-4x)}ln(3))}{3^{(7x)}}\end{align*}$

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Calculus 2 : First and Second Derivatives of Functions

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1566 : Calculus Ii

Example Question #441 : Derivatives

Example Question #441 : Derivative Review

Example Question #1562 : Calculus Ii

Example Question #441 : Derivatives

Example Question #241 : First And Second Derivatives Of Functions

Example Question #442 : Derivative Review

Example Question #241 : First And Second Derivatives Of Functions

Example Question #444 : Derivative Review

Example Question #445 : Derivative Review

All Calculus 2 Resources

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