Finding Derivatives Made Easy! Derivative of A Chain of Functions
[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-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"]
Definition:
The chain rule is a method in calculus used to find the derivative of a composite function. It’s like a two-step process: first, you take the derivative of the outer function, and then you multiply it by the derivative of the inner function. This helps you understand how changes in one variable affect a chain of functions. For function \( f(x) \) and \( g(x) \), to find the derivative of \( f(g(x)) \), we have: For additional educational resources,.
\( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \)
Let’s consider this example: \( \text{Find the derivative of } (x + 1)^3. \)
- Define the inner and outer functions and their derivatives
\( g(x) = x + 1, \ g'(x) = 1 \)
\( f(u) = u^3, \ f'(u) = 3u^2 \)
- Apply the chain rule
\( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) = 3(x + 1)^2 \cdot 1 \)
Chain rule for more functions:
Differentiating a composition of four functions using the chain rule involves taking the derivative of each function sequentially and multiplying them together. It’s like unwrapping nested functions layer by layer, applying derivatives at each step to reveal the rate of change of the entire composition.
\( f = f(u), \ u = u(v), \ v = v(w), \ w = w(x) \)
\( \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx} \)
This represents the extended chain rule formula for the case where you have a composition of four functions, allowing you to differentiate \( f(u(v(w(x)))) \) with respect to \( x \).
Here is an example for this formula: \( \text{Find the derivative of } \sin(\cos(x^3)). \)
1.Define functions
\( v(x) = x^3 \)
\( u(v) = \cos v \)
\( f(u) = \sin u \)
2. Define derivatives
\( v'(x) = 3x^2 \)
\( u'(v) = -\sin v \)
\( f'(u) = \cos u \)
3. Apply chain rule
\( \frac{df}{dx} = f'(u) \cdot u'(v) \cdot v'(x) \)
4. Simplify
\( = \cos(\cos(x^3)) \cdot (-\sin(x^3)) \cdot 3x^2 \)
This next example includes radical and logarithm derivatives, which will be explained later. But if you already know how the derivative of those functions work, then this problem is for you to better understand the chain rule:
\( \text{Find the derivative of } \sqrt{\ln(e^{x^2})}. \)
1.Define functions
\( w(x) = x^2 \)
\( v(w) = e^w \)
\( u(v) = \ln(v) \)
\( f(u) = \sqrt{u} \)
2. Define derivatives
\( w'(x) = 2x \)
\( v'(w) = e^w \)
\( u'(v) = \frac{1}{v} \)
\( f'(u) = \frac{1}{2\sqrt{u}} \)
3. Apply chain rule
\( \frac{df}{dx} = f'(u) \cdot u'(v) \cdot v'(w) \cdot w'(x) \)
4. Simplify
\( = \frac{1}{2\sqrt{\ln(e^{x^2})}} \cdot \frac{1}{e^{x^2}} \cdot e^{x^2} \cdot 2x \)
Related to This Article
More math articles
- The Enchanted Forest of How to Compare Ratios – A Tale of Mathematical Adventure
- Top 10 ISEE Upper-Level Math Practice Questions
- 6th Grade Math Worksheets: FREE & Printable
- The Ultimate 7th Grade OAA Math Course (+FREE Worksheets)
- Zero and One: The Fundamental Pillars of Mathematics
- The Ultimate MCA Algebra 1 Course (+FREE Worksheets)
- How to Help Your 3rd Grade Student Prepare for the District of Columbia DC CAPE Math Test
- 6th Grade OAA Math Worksheets: FREE & Printable
- HiSET Testing Accommodations for Students with Disabilities
- How to Understand the Nuance of Equality of Vectors in Two Dimensions
What people say about "Finding Derivatives Made Easy! Derivative of A Chain of Functions - Effortless Math: We Help Students Learn to LOVE Mathematics"?
No one replied yet.