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
- The Ultimate 7th Grade PARCC Math Course (+FREE Worksheets)
- The Remainder Theorem
- How to Compare Decimals? (+FREE Worksheet!)
- The Ultimate Precalculus Course (+FREE Worksheets & Tests)
- Top 10 Tips to ACE the CBEST Math Test
- How to Solve Trigonometric Equations?
- 4th Grade FSA Math Practice Test Questions
- Full-Length DAT Quantitative Reasoning Practice Test-Answers and Explanations
- 6th Grade SBAC Math Worksheets: FREE & Printable
- How to Decode Decimals: Unveiling the Value of Each Digit
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.