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Derivatives of Trigonometric Functions 1Derivatives of Trigonometric Functions 1
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Question 1 of 6
1. Question
Find the derivative using the chain rule`5cos4x`Hint
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Derivatives of Trigonometric Functions
`y'(\text(sin))=\text(cos)``y'(\text(cos))=-\text(sin)``y'(\text(tan))=\text(sec)^2`Chain Rule
$$y’=\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$First, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$5\;\text{cos}\;\color{#9a00c7}{4x}$$ `x` `=` `4x` `n` `=` `1` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$5\cdot\color{#e65021}{1}\cdot\text{cos}\;(\color{#004ec4}{4x})\cdot\color{#00880A}{f'(4x)}$$ Substitute known values `=` $$5\cdot(-\text{sin}\;4x)\cdot\color{#00880A}{4}$$ Differentiate the values `=` $$-20\;\text{sin}\;4x$$ `y’=-20 \text(sin) 4x` -
Question 2 of 6
2. Question
Find the derivative using the chain rule`tan(4x-2)`Hint
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Derivatives of Trigonometric Functions
`y'(\text(sin))=\text(cos)``y'(\text(cos))=-\text(sin)``y'(\text(tan))=\text(sec)^2`Chain Rule
$$y’=\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$First, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$\color{#9a00c7}{\text{tan}\;(4x-2)}$$ `x` `=` `(4x-2)` `n` `=` `1` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{1}\cdot\text{tan}\;(\color{#004ec4}{4x-2})\cdot\color{#00880A}{f'(4x-2)}$$ Substitute known values `=` $$(\text{sec}^2\;(4x-2)\cdot\color{#00880A}{4}$$ Differentiate the values `=` $$4\;\text{sec}^2\;(4x-2)$$ Simplify `y’=4 \text(sec)^2 (4x-2)` -
Question 3 of 6
3. Question
Find the derivative using the chain rule`sqrt(cosx)`Hint
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Derivatives of Trigonometric Functions
`y'(\text(sin))=\text(cos)``y'(\text(cos))=-\text(sin)``y'(\text(tan))=\text(sec)^2`Chain Rule
$$y’=\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$First, convert the surd into an exponent`sqrt(\text(cos)x)` `=` `\text(cos) x^(1/2)` Next, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$\color{#9a00c7}{\text{cos}\;x}^{\color{#e65021}{\frac{1}{2}}}$$ `x` `=` `\text(cos) x` `n` `=` `1/2` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{\frac{1}{2}}\cdot(\color{#004ec4}{\text{cos}\;x})^{\color{#e65021}{(\frac{1}{2})}-1}\cdot\color{#00880A}{f'(\text{cos}\;x)}$$ Substitute known values `=` $$\frac{1}{2}(\text{cos}\;x)^{-\frac{1}{2}}\cdot\color{#00880A}{-\text{sin}\;x}$$ Differentiate `\text(cos) x` `=` $$\frac{-\text{sin}\;x}{2(\text{cos}\;x)^{\frac{1}{2}}}$$ Reciprocate `\text(cos) x^(-1/2)` `=` $$\frac{-\text{sin}\;x}{2\sqrt{\text{cos}\;x}}$$ Convert the exponent into a surd `y’=(-\text(sin) x)/(2sqrt(\text(cos) x))` -
Question 4 of 6
4. Question
Find the derivative using the chain rule`sin^4x`Hint
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Derivatives of Trigonometric Functions
`y'(\text(sin))=\text(cos)``y'(\text(cos))=-\text(sin)``y'(\text(tan))=\text(sec)^2`Chain Rule
$$y’=\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$The expression can also be written as`(\text(sin) x)^4` Next, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$\color{#9a00c7}{\text{sin}\;x}^{\color{#e65021}{4}}$$ `x` `=` `\text(sin) x` `n` `=` `4` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{4}\cdot(\color{#004ec4}{\text{sin}\;x})^{\color{#e65021}{4}-1}\cdot\color{#00880A}{f'(\text{sin}\;x)}$$ Substitute known values `=` $$4\;(\text{sin}\;x)^{3}\cdot\color{#00880A}{\text{cos}\;x}$$ Differentiate the values `=` $$4\;\text{sin}^3\;x\;\text{cos}\;x$$ Evaluate `y’=4 \text(sin)^3 x \text(cos) x` -
Question 5 of 6
5. Question
Find the derivative using the chain rule`cos^3 2x`Hint
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Excellent!
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Derivatives of Trigonometric Functions
`y'(\text(sin))=\text(cos)``y'(\text(cos))=-\text(sin)``y'(\text(tan))=\text(sec)^2`Chain Rule
$$y’=\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$The expression can also be written as`(\text(cos) 2x)^3` Next, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$\color{#9a00c7}{\text{cos}\;2x}^{\color{#e65021}{3}}$$ `x` `=` `\text(cos) 2x` `n` `=` `3` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{3}\cdot(\color{#004ec4}{\text{cos}\;2x})^{\color{#e65021}{3}-1}\cdot\color{#00880A}{f'(\text{cos}\;2x)}$$ Substitute known values `=` $$3\;(\text{cos}\;2x)^{2}\cdot\color{#00880A}{-\text{sin}\;2x\cdot2}$$ Differentiate the values `=` $$-6\;\text{cos}^2\;2x\;\text{sin}\;2x$$ Evaluate `y’=-6 \text(cos)^2 2x \text(sin) 2x` -
Question 6 of 6
6. Question
Find the derivative using the chain rule`3 sin(5x-3)`Hint
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Nice Job!
Incorrect
Derivatives of Trigonometric Functions
`y'(\text(sin))=\text(cos)``y'(\text(cos))=-\text(sin)``y'(\text(tan))=\text(sec)^2`Chain Rule
$$y’=\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$First, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$3\;\text{sin}\;\color{#9a00c7}{(5x-3)}$$ `x` `=` `5x-3` `n` `=` `1` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$3\cdot\color{#e65021}{1}\cdot\text{sin}\;(\color{#004ec4}{5x-3})\cdot\color{#00880A}{f'(5x-3)}$$ Substitute known values `=` $$3\;(\text{cos}\;(5x-3))\cdot5$$ Differentiate the values `=` $$15\;\text{cos}\;(5x-3)$$ Evaluate `y’=15 \text(cos)(5x-3)`
Quizzes
- Power Rule 1
- Power Rule 2
- Power Rule 3
- Power Rule 4
- Chain Rule 1
- Chain Rule 2
- Product Rule
- Quotient Rule
- Derivatives of Exponential Functions 1
- Derivatives of Exponential Functions 2
- Derivatives of Exponential Functions 3
- Derivatives of Trigonometric Functions 1
- Derivatives of Trigonometric Functions 2
- Derivatives of Trigonometric Functions 3