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Question 1 of 5
Find the derivative using the chain rule
(3x+tan 7x)6
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Derivatives of Trigonometric Functions
Chain Rule
y′=n⋅(f(x))n−1⋅f′(x)
First, identify the values of the function
| f(x) |
= |
xn |
| f(x) |
= |
(3x+tan7x)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)
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Question 2 of 5
Find the derivative using the chain rule
sin^(-1)x
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)}
The expression can also be written as
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}} |
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)
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Question 3 of 5
Find the derivative using the chain rule
(cos x+sin x)^2
Incorrect
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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)
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Question 4 of 5
Find the derivative using the chain rule
3/(sin^2 3x)
Incorrect
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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
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}} |
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)
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Question 5 of 5
Find the derivative using the product rule
x^4 cos3x
Incorrect
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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
