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Derivative of a Trigonometric Function>
Derivative of a Trigonometric Function 3Derivative of a Trigonometric Function 3
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Question 1 of 5
1. Question
Find the derivative using the product rule`tan3x 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`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` `=` `\text(tan) 3x` `u’` `=` `(du)/(dx)` `=` `3 sec^2 3x` Product Rule Derivative of `v`:`v` `=` `\text(cos) x` `v’` `=` `(dv)/(dx)` `=` `-\text(sin) x` 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)` `=` $$(\color{#9a00c7}{\text{cos}\;x}\cdot\color{#e65021}{3\;sec^2\;3x})+[\color{#00880A}{\text{tan}\;3x}\cdot(\color{#004ec4}{-\text{sin}\;x})]$$ Substitute known values `y’` `=` `3\text(cos) 3x \text(sec)^2 3x-\text(sin) x \text(tan) 3x` Evaluate `y’=3\text(cos) 3x \text(sec)^2 3x-\text(sin) x \text(tan) 3x` -
Question 2 of 5
2. Question
Find the derivative using the product rule`8x sin(x/4)`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` `=` `8x` `u’` `=` `(du)/(dx)` `=` `8` Power Rule Derivative of `v`:`v` `=` `\text(sin) x/4` `v’` `=` `(dv)/(dx)` `=` `1/4\text(cos) x/4` Power 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)` `=` $$\left(\color{#9a00c7}{\text{sin}\;\frac{x}{4}}\cdot\color{#e65021}{8}\right)+\left(\color{#00880A}{8x}\cdot\color{#004ec4}{\frac{1}{4}}\text{cos}\;\frac{x}{4}\right)$$ Substitute known values `y’` `=` `8\text(sin) x/4+2x\text(cos) x/4` Evaluate `y’=8\text(sin) x/4+2x\text(cos) x/4` -
Question 3 of 5
3. Question
Find the derivative using the product rule`x^2 cos^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`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^2` `u’` `=` `(du)/(dx)` `=` `2x` Power Rule Derivative of `v`:`v` `=` `(\text(cos) x)^3` `v’` `=` `(dv)/(dx)` `=` `-3\text(cos)^2 x \text(sin) x` 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)` `=` $$(\color{#9a00c7}{\text{cos}^3\;x}\cdot\color{#e65021}{2x})+(\color{#00880A}{x^2}\cdot\color{#004ec4}{-3\text{cos}^2x\;\text{sin}\;x})$$ Substitute known values `y’` `=` `2x\text(cos)^3 x-3x^2\text(cos)^2x \text(sin) x` Evaluate `y’=2x\text(cos)^3 x-3x^2\text(cos)^2x \text(sin) x` -
Question 4 of 5
4. Question
Find the derivative using the quotient rule`(tanx)/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`Quotient Rule
$$\frac{dy}{dx}=\frac{\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}-\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}}{\color{#9a00c7}{v}^2}$$First, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `\text(tan) x` `u’` `=` `(du)/(dx)` `=` `\text(sec)^2 x` Derivative of `v`:`v` `=` `x` `v’` `=` `(dv)/(dx)` `=` `1` Power Rule Substitute the components into the quotient rule$$\frac{dy}{dx}$$ `=` $$\frac{\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}-\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}}{\color{#9a00c7}{v}^2}$$ `(dy)/(dx)` `=` $$\frac{(\color{#9a00c7}{x}\cdot\color{#e65021}{\text{sec}^2\;x})-(\color{#00880A}{\text{tan}\;x}\cdot\color{#004ec4}{1})}{\color{#9a00c7}{x}^2}$$ Substitute known values `y’` `=` `(x \text(sec)^2 x-\text(tan) x)/(x^2)` Evaluate `y’=(x \text(sec)^2 x-\text(tan) x)/(x^2)` -
Question 5 of 5
5. Question
Find the derivative using the quotient rule`(sin3x)/(cos4x)`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`Quotient Rule
$$\frac{dy}{dx}=\frac{\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}-\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}}{\color{#9a00c7}{v}^2}$$First, find the derivative of `u` and `v`Derivative of `u`:`u` `=` `\text(sin) 3x` `u’` `=` `(du)/(dx)` `=` `3\text(cos) 3x` Chain Rule Derivative of `v`:`v` `=` `\text(cos) 4x` `v’` `=` `(dv)/(dx)` `=` `-4\text(sin) 4x` Chain Rule Substitute the components into the quotient rule$$\frac{dy}{dx}$$ `=` $$\frac{\color{#9a00c7}{v}\color{#e65021}{\frac{du}{dx}}-\color{#00880A}{u}\color{#004ec4}{\frac{dv}{dx}}}{\color{#9a00c7}{v}^2}$$ `(dy)/(dx)` `=` $$\frac{(\color{#9a00c7}{\text{cos}\;4x}\cdot\color{#e65021}{3\text{cos}\;3x})-[\color{#00880A}{\text{sin}\;3x}\cdot(\color{#004ec4}{-4\text{sin}\;4x})]}{\color{#9a00c7}{\text{cos}\;4x}^2}$$ Substitute known values `y’` `=` `(3\text(cos) 4x \text(cos)3x+4\text(sin) 3x \text(sin) 4x)/((\text(cos)4x)^2)` Evaluate `y’=(3\text(cos) 4x \text(cos)3x+4\text(sin) 3x \text(sin) 4x)/((\text(cos)4x)^2)`
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- Converting Angle Measures 2
- Converting Angle Measures 3
- Finding the Central Angle in a Circle
- Finding Areas in a Circle
- Values on the Unit Circle
- Finding Missing Angles Using the Unit Circle
- Trigonometric Ratios in the Unit Circle
- Trig Exact Values 1
- Trig Exact Values 2
- Trig Equations
- Derivative of a Trigonometric Function 1
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- Derivative of a Trigonometric Function 3
- Applications of Differentiation
- Integral of a Trigonometric Function 1
- Integral of a Trigonometric Function 2
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- Graphing Trigonometric Functions 3
- Graphing Trigonometric Functions 4