How to Find The Derivative of a Trigonometric Function
Using the limit formula to derive functions, we can uncover the general derivative form of trigonometric functions. This process requires an understanding of trigonometric identities. By applying these identities within the limit framework, it becomes possible to systematically determine the derivatives of various trigonometric functions, enhancing calculus applications.

Derivative of \( sin(x) \) :
\( \begin{align*}
\frac{d}{dx} \sin x &= \lim_{h \to 0} \frac{\sin(x + h) – \sin x}{h} \\
&= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h – \sin x}{h} \\
&= \lim_{h \to 0} \left(\sin x \cdot \frac{\cos h – 1}{h} + \cos x \cdot \frac{\sin h}{h}\right) \\
&= \sin x \cdot \lim_{h \to 0} \frac{\cos h – 1}{h} + \cos x \cdot \lim_{h \to 0} \frac{\sin h}{h} \\
&= \sin x \cdot 0 + \cos x \cdot 1 \\
&= \cos x
\end{align*}\)
So the derivative of \(sin (x)\) , is \(cos (x)\). Now, using chain rule, we can have:
\( sin(f(x))=cos (f(x))×f’ (x) \)
Derivative of \(cos (x) \):
Almost the same procedure can be used to find the derivative of \( cos (x) \) :
\(\begin{align*}
\frac{d}{dx} \cos x &= \lim_{h \to 0} \frac{\cos(x + h) – \cos x}{h} \\
&= \lim_{h \to 0} \frac{\cos x \cos h – \sin x \sin h – \cos x}{h} \\
&= \lim_{h \to 0} \left(\cos x \cdot \frac{\cos h – 1}{h} – \sin x \cdot \frac{\sin h}{h}\right) \\
&= \cos x \cdot \lim_{h \to 0} \frac{\cos h – 1}{h} – \sin x \cdot \lim_{h \to 0} \frac{\sin h}{h} \\
&= \cos x \cdot 0 – \sin x \cdot 1 \\
&= -\sin x
\end{align*}\)
Derivative of \(tan (x) \) and \(cot (x) \):
Now we could do the same thing for tangent and cotangent, but it would be easier to use quotient rule and find them using \( \tan x = \frac{\sin x}{\cos x} \) and \( \cot x = \frac{\cos x}{\sin x} \):
\( \frac{d}{dx} \tan x = \frac{d}{dx} \left( \frac{\sin x}{\cos x} \right) = \frac{\cos x \cdot \cos x – \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x \)
And:
\( \frac{d}{dx} \cot x = \frac{d}{dx} \left( \frac{\cos x}{\sin x} \right) = \frac{\sin x \cdot (-\sin x) – \cos x \cdot \cos x}{\sin^2 x} = \frac{-\sin^2 x – \cos^2 x}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x \)
Related to This Article
More math articles
- How to Calculate the Area, Perimeter, and Radius of Quarter Circles
- FREE 4th Grade Common Core Math Practice Test
- How to Add and Subtract Polynomials Using Algebra Tiles
- Top 10 7th Grade MEAP Math Practice Questions
- How to Master Two-Column Proofs: A Step-by-Step Tutorial
- 6th Grade Georgia Milestones Assessment System Math Worksheets: FREE & Printable
- HSPT Math FREE Sample Practice Questions
- What is the Relationship between Dilations and Angles in Geometry
- Optimizing Betting Strategies in Crash Gambling with Mathematics
- Play the Math Game: How to Craft Tables and Graphs for Two-variable Equations
What people say about "How to Find The Derivative of a Trigonometric Function - Effortless Math: We Help Students Learn to LOVE Mathematics"?
No one replied yet.