12 Questions
What is the derivative of the tangent function, ( \tan(x) )?
( \sec^2(x) )
What is the derivative of the arccosine function?
( -\frac{1}{\sqrt{1-y^2}} )
Why is finding the derivative of ( \tan(x) ) more complex compared to other trigonometric functions?
It includes fractions that need simplification
What is the derivative of ( \arcsin(y) )?
( \frac{1}{\sqrt{1-y^2}} )
What is the derivative of the cotangent function, ( \cot(x) )?
( -\csc^2(x) )
How do trigonometric functions transform when rewritten using inverse operations?
Their derivatives transform based on the operation used
What is the derivative of $\sin(x)$?
$\cos(x)$
If $y = \sin(x)$, what is the slope of the graph at $x = \frac{\pi}{3}$?
$\frac{1}{2}$
What is the derivative of $\cos(x)$?
$-\sin(x)$
At what angle does $\cos(x)$ reach its maximum value?
$0$
If $y = \sin(x)$, what is the derivative with respect to $x$ at $x = \frac{\pi}{6}$?
$\frac{1}{2}$
What is the negative of the derivative of $\cos(x)$?
$-\sin(x)$
Study Notes
Derivatives of Trigonometric Functions
In calculus, we often need to find derivatives—the rates of change—of various mathematical expressions. When it comes to working with trigonometric functions like sine, cosine, tangent, and their inverse counterparts, understanding how they behave analytically is crucial. Here, I'll guide you through these fundamentals so you can confidently compute derivatives of common trigonometric functions.
Sine Function
The derivative of ( \sin(x) ) is given by ( \cos(x) ):
[ \frac{d}{dx} (\sin x) = \cos x ]
This means that if ( y=\sin x ), the slope of the graph at any point will equal the cosine value there. For instance, if ( x=60^\circ ), the corresponding values would be ( y=\sin 60^\circ =\frac{\sqrt{3}}{2} ) and its slope ( dy/dx=\cos 60^\circ=\frac{1}{2} ).
Cosine Function
Similarly, the derivative of ( \cos(x) ) is ( -\sin(x) ):
[ \frac{d}{dx} (\cos x) = - \sin x ]
To visualize this, imagine a circle rolling along the horizontal axis; the angle between its radius vector and positive direction changes constantly. As the angle increases from 0 to ( \pi ), both ( \cos x ) and ( -\sin x ) decrease following the negative slope of the cosine function curve.
Tangent Function
The derivative of ( \tan(x)=\frac{\sin(x)}{\cos(x)} ) is more complex because it involves fractions, which must be simplified before finding the derivative itself:
[ \begin{align*} & \frac{d(\tan x)}{dx}=\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)\ &\quad = \frac{(\cos x)(d(\sin x)/dx)-(\sin x)(d(\cos x)/dx)}{(\cos^2 x)}\ &\quad = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}\ &\quad = 1+\cot^2 x \ &\quad = \sec^2 x \end{align*} ]
Therefore, ( \frac{d}{dx}(\tan x) = \sec^2 x ). In other words, the rate of change of the tangent of ( x ) equals the square of secant of ( x ).
Other Identities
Arcsin, arccos, and arctan have derivatives of ( \frac{1}{\sqrt{1-(y)^2}} ), ( -\frac{1}{\sqrt{(1-(y)^2)}} ), and ( \frac{1}{(1+(y)^2)} ), respectively, where ( y ) represents each respective function argument inside the inverse operation. These formulas signify that when rewriting our original trigonometric functions using inverse operations, their derivatives transform accordingly.
Being able to differentiate trigonometric functions allows us to solve problems involving optimization, kinematics, and many other areas requiring an understanding of how quantities vary over time. So, mastering these techniques will serve as a valuable tool for your future endeavors in mathematics!
Learn how to find derivatives of trigonometric functions such as sine, cosine, and tangent. Explore the relationships between these functions and their derivatives, and understand how to compute them analytically. Mastering these derivative rules is essential for solving calculus problems involving trigonometric functions.
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