Limits and Imp/Explicit Differentiation

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SUMMARY

The discussion focuses on the differentiability of the piecewise function F(x) defined as: F(x) = -x for x < -1, F(x) = x² for -1 ≤ x < 1, and F(x) = 1/x for x ≥ 1. The limit definition of the derivative was applied to check differentiability at x = -1 and x = 1, concluding that the function is not differentiable at both points. Additionally, the slope of the curve defined by y = (t^(2/3))e^(sin(t)) + t at the point (0,0) was calculated, resulting in a derivative of 1.

PREREQUISITES
  • Understanding of piecewise functions
  • Knowledge of the limit definition of the derivative
  • Familiarity with basic calculus concepts, including differentiation
  • Experience with exponential and trigonometric functions
NEXT STEPS
  • Study the properties of piecewise functions in calculus
  • Learn about the limit definition of the derivative in depth
  • Explore differentiability conditions for functions
  • Investigate the application of derivatives in real-world scenarios
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Students preparing for calculus exams, educators teaching differentiation, and anyone seeking to deepen their understanding of limits and derivatives in mathematical analysis.

Sesse
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*opps fixing the latex code*
I have a test coming up so I tried solving some review questions but couldn't solve all of them. I will also post the ones which I am not sure whether they are correct or wrong so please spare some time to check them too. :smile:
1.
[tex]F(x)=<br /> \left{<br /> \begin{array}{cc}<br /> -x, & \mbox{ if } x<-1\\<br /> x^2, & \mbox{ if } -1 \leq x<1\\<br /> \frac{1}{x}, & \mbox{ if } x\geq 1<br /> \end{array}<br /> \right[/tex]
Use limit definition of the derivative to see if the fuction is differentiable at [tex]x=-1[/tex] and [tex]x=1[/tex]
At [tex]x=-1[/tex]
[tex] \lim_{x\rightarrow -1^-} \frac{f(x)-f(-1)}{x-(-1)}=-1[/tex]
[tex]\lim_{x\rightarrow -1^+} \frac{f(x)+1}{x+1} = \frac{(x+1)(x-1)}{x+1}=x-1=-2[/tex]
If this solution for [tex]x=-1[/tex] is correct then it is also correct for [tex]x=1[/tex] so I won't show my solution for that.
So the fuction is not differentiable at both points?
2. Probably correct.
3. [tex]y=(t^{2/3}).e^{sin(t)}+t[/tex]
Find the slope at (0,0)
[tex]\frac{dy}{dx}=\left(\frac{2}{3}t^{-1/3}e^{sin(t)}+t^{2/3}e^{sin(t)}cos(t)\right)+1[/tex]
[tex]\frac{dy}{dx}=1?[/tex]

More will come. :)
 
Last edited:
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Yes, both of those are correct.
 

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