Essentials Calculus: Comparing Early Transcendentals

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The discussion clarifies the differences between "Essentials Calculus" and "Essentials Calculus: Early Transcendentals." The key distinction lies in the order of topics, particularly regarding the treatment of trigonometric and exponential functions. The Early Transcendentals edition includes these functions earlier in the course, while the standard version assumes prior knowledge. Additionally, there are full and single-variable versions of the Essentials Calculus texts, with the latter covering fewer chapters. It's essential to confirm which specific edition is required for the course before making a purchase.
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"Essentials Calculus"

Hello,

I just registered for a calculus course and the math department told me to buy "Stewart's 'Essentials Calculus' book".

Easy enough to find online, but I also see there is another published called "Essentials Calculus: Early Transcendentals"



Anyone know the difference between these books? the description on amazon etc is the same for both, but the price is different (slightly) and obviously the title.
 
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Same topics, but covered in a different order.

So get the edition suggested for your course. If "Early Transcendentals" was not mentioned by them, do not get that version.
 


thank you kindly!
 


The various editions of Stewart's Calculus texts are described at:

http://www.stewartcalculus.com/

The Early Transcendentals Essential Calculus has a blue cover while the other has a red one. The principal difference as already mentioned is the order of topics - specifically whether trigonometric and exponential functions (which are transcendental functions) are to be presented as part of the course content or are they assumed knowledge (hence "early" trascendentals).

On top of this, the Essential Calculus texts come in full and "Single Variable" versions. The latter only goes up to Chapter 11 (I think) while the former covers all chapters and includes the mutlivaraite material (usually covered) in Calculus III.

As mentioned, I'd make very sure which text is intended before ordering/purchasing.

--Elucidus
 
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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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