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.
[include_netrun_products_block from-products="product/6-south-carolina-sc-ready-grade-3-math-practice-tests/" product-list-class="bundle-products float-left" product-item-class="float-left" product-item-image-container-class="p-0 float-left" product-item-image-container-size="col-2" product-item-image-container-custom-style="" product-item-container-size="" product-item-add-to-cart-class="btn-accent btn-purchase-ajax" product-item-button-custom-url="{url}/?ajax-add-to-cart={id}" product-item-button-custom-url-if-not-salable="{productUrl} product-item-container-class="" product-item-element-order="image,title,purchase,price" product-item-title-size="" product-item-title-wrapper-size="col-10" product-item-title-tag="h3" product-item-title-class="mt-0" product-item-title-wrapper-class="float-left pr-0" product-item-price-size="" product-item-purchase-size="" product-item-purchase-wrapper-size="" product-item-price-wrapper-class="pr-0 float-left" product-item-price-wrapper-size="col-10" product-item-read-more-text="" product-item-add-to-cart-text="" product-item-add-to-cart-custom-attribute="title='Purchase this book with single click'" product-item-thumbnail-size="290-380" show-details="false" show-excerpt="false" paginate="false" lazy-load="true"] [include_netrun_products_block from-products="product/6-ohio-ost-grade-3-math-practice-tests/" product-list-class="bundle-products float-left" product-item-class="float-left" product-item-image-container-class="p-0 float-left" product-item-image-container-size="col-2" product-item-image-container-custom-style="" product-item-container-size="" product-item-add-to-cart-class="btn-accent btn-purchase-ajax" product-item-button-custom-url="{url}/?ajax-add-to-cart={id}" product-item-button-custom-url-if-not-salable="{productUrl} product-item-container-class="" product-item-element-order="image,title,purchase,price" product-item-title-size="" product-item-title-wrapper-size="col-10" product-item-title-tag="h3" product-item-title-class="mt-0" product-item-title-wrapper-class="float-left pr-0" product-item-price-size="" product-item-purchase-size="" product-item-purchase-wrapper-size="" product-item-price-wrapper-class="pr-0 float-left" product-item-price-wrapper-size="col-10" product-item-read-more-text="" product-item-add-to-cart-text="" product-item-add-to-cart-custom-attribute="title='Purchase this book with single click'" product-item-thumbnail-size="290-380" show-details="false" show-excerpt="false" paginate="false" lazy-load="true"]
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 Simplify Fractions? (+FREE Worksheet!)
- 10 Tips for Advanced Studying Mathematics
- Linear, Quadratic, and Exponential Models
- Bеѕt Cоllеgе Lарtорs in 2026
- Mastering Translations on the Coordinate Plane: A Step-by-Step
- What Kind of Math Is on the Computer Science?
- How to Use Models to Decompose Fractions into Unit Fractions?
- Your Winning Game Plan: How to Use Angle Relationships to Write and Solve Equations
- 10 Most Common 3rd Grade MCAS Math Questions
- Full-Length Accuplacer Math Practice Test
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.