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Derivatives of Trigonometric Functions 2Derivatives of Trigonometric Functions 2
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Question 1 of 5
1. Question
Find the derivative using the chain rule`(3x+tan 7x)^6`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)` `=` $$\color{#9a00c7}{(3x+\text{tan}\;7x)}^{\color{#e65021}{6}}$$ `x` `=` `3x+\text(tan) 7x` `n` `=` `6` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{6}\cdot(\color{#004ec4}{3x+\text{tan}\;7x})^{\color{#e65021}{6}-1}\cdot\color{#00880A}{f'(3x+\text{tan}\;7x)}$$ Substitute known values `=` $$6\;(3x+\text{tan}\;7x)^{5}\cdot\color{#00880A}{3+\text{sec}^2\;7x\cdot7}$$ Differentiate the values `=` $$6\;(3x+\text{tan}\;7x)^{5}(3+7\text{sec}^2\;7x)$$ Evaluate `y’=6(3x+\text(tan) 7x)^5(3+7\text(sec)^2 7x)` -
Question 2 of 5
2. Question
Find the derivative using the chain rule`sin^(-1)x`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)^(-1)` Next, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$\color{#9a00c7}{\text{sin}\;x}^{\color{#e65021}{-1}}$$ `x` `=` `\text(sin) x` `n` `=` `-1` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{-1}\cdot(\color{#004ec4}{\text{sin}\;x})^{\color{#e65021}{-1}-1}\cdot\color{#00880A}{f'(\text{sin}\;x)}$$ Substitute known values `=` $$(-\text{sin}\;x)^{-2}\cdot\color{#00880A}{\text{cos}\;x}$$ Differentiate the values `=` $$-\frac{\text{cos}\;x}{\text{sin}^2\;x}$$ Reciprocate `\text(sin)^2x` `y’=-(\text(cos) x)/(\text(sin)^2 x)` -
Question 3 of 5
3. Question
Find the derivative using the chain rule`(cos x+sin x)^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)` `=` $$\color{#9a00c7}{(\text{cos}\;x+\text{sin}\;x)}^{\color{#e65021}{2}}$$ `x` `=` `\text(cos) x+\text(sin) x` `n` `=` `2` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{2}\cdot(\color{#004ec4}{\text{cos}\;x+\text{sin}\;x})^{\color{#e65021}{2}-1}\cdot\color{#00880A}{f'(\text{cos}\;x+\text{sin}\;x)}$$ Substitute known values `=` $$2\;(\text{cos}\;x+\text{sin}\;x)(\color{#00880A}{-\text{sin}\;x+\text{cos}\;x})$$ Differentiate the values `=` $$2(\text{cos}^2\;x-\text{sin}^2\;x)$$ Evaluate `y’=2(\text(cos)^2 x-\text(sin)^2 x)` -
Question 4 of 5
4. Question
Find the derivative using the chain rule`3/(sin^2 3x)`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`3/(\text(sin) 3x)^2` Remove the denominator by reciprocating its value`3/(\text(sin) 3x)^2` `=` `3(\text(sin) 3x)^(-2)` Reciprocate the denominator Next, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$3(\color{#9a00c7}{\text{sin}\;3x})^{\color{#e65021}{-2}}$$ `x` `=` `\text(sin) 3x` `n` `=` `-2` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$3\cdot\color{#e65021}{-2}\cdot(\color{#004ec4}{\text{sin}\;3x})^{\color{#e65021}{-2}-1}\cdot\color{#00880A}{f'(\text{sin}\;3x)}$$ Substitute known values `=` $$-6\;(\text{sin}\;3x)^{-3}\cdot\color{#00880A}{\text{cos}\;3x\cdot3}$$ Differentiate the values `=` $$\frac{-6\cdot3\text{cos}\;3x}{\text{sin}^3\;3x}$$ Reciprocate `(\text(sin) 3x)^(-3)` `=` $$\frac{-18\;\text{cos}\;3x}{\text{sin}^3\;3x}$$ Evaluate `y’=(-18 \text(cos) 3x)/(\text(sin)^3 3x)` -
Question 5 of 5
5. Question
Find the derivative using the product rule`x^4 cos3x`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`Product Rule
$$\frac{dy}{dx}=\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$$First, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `x^4` `u’` `=` `(du)/(dx)` `=` `4x^3` Power Rule Derivative of `v`:`v` `=` `\text(cos) 3x` `v’` `=` `(dv)/(dx)` `=` `-\text(sin) 3x*3` Chain Rule Substitute the components into the product rule$$\frac{dy}{dx}$$ `=` $$\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}+\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}$$ `(dy)/(dx)` `=` $$(\color{#9a00c7}{\text{cos}\;3x}\cdot\color{#e65021}{4x^3})+(\color{#00880A}{x^4}\cdot\color{#004ec4}{-\text{sin}\;3x\cdot3})$$ Substitute known values `y’` `=` `4x^3\text(cos) 3x-3x^4\text(sin) 3x` Evaluate `y’=4x^3\text(cos) 3x-3x^4\text(sin) 3x`
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