- #1
Gwapako
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Help with Derivatives Problem - Get Answers Now
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SUMMARY
The discussion focuses on solving derivatives using the limit definition, specifically applying the formula $\displaystyle f’(x) = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x}$. Participants detail the algebraic steps necessary to simplify expressions, including the use of common denominators and rationalizing numerators. The limits for both polynomial and radical functions are addressed, emphasizing the importance of limit properties in derivative calculations.
PREREQUISITES- Understanding of calculus concepts, particularly limits and derivatives.
- Familiarity with algebraic manipulation, including combining fractions and rationalizing expressions.
- Knowledge of polynomial and radical functions.
- Basic proficiency in mathematical notation and limit properties.
- Study the formal definition of derivatives in calculus.
- Practice algebraic techniques for simplifying limits, including finding common denominators.
- Explore the application of the limit definition to various types of functions, including trigonometric and exponential functions.
- Learn about higher-order derivatives and their significance in calculus.
Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of derivative calculations and limit applications.
- #2
skeeter
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I assume you’re referring to a form of the following limit ...
$\displaystyle f’(x) = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x}$
for ease in typing it out, let $\Delta x = h$
$\displaystyle y’ = \lim_{h \to 0} \dfrac{1}{h} \left[\dfrac{2(x+h)-1}{2(x+h)+1} - \dfrac{2x-1}{2x+1} \right]$
from here, it’s just the algebra drill of combining the two fractions using a common denominator and ultimately getting the $h$ in the leading $\dfrac{1}{h}$ factor to divide out with an $h$ factor in the numerator.by the same token, let $\Delta t = h$ in the second limit; also, using the properties of limits will make the overall task a bit easier ...
$\displaystyle x’ = 3 \cdot \lim_{h \to 0} \dfrac{(t+h)^2-t^2}{h} - 2 \cdot \lim_{h \to 0} \dfrac{\sqrt{t+h} - \sqrt{t}}{h}$
the limit of the first term is straightforward; the second will require you to rationalize the numerator
$\displaystyle f’(x) = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x}$
for ease in typing it out, let $\Delta x = h$
$\displaystyle y’ = \lim_{h \to 0} \dfrac{1}{h} \left[\dfrac{2(x+h)-1}{2(x+h)+1} - \dfrac{2x-1}{2x+1} \right]$
from here, it’s just the algebra drill of combining the two fractions using a common denominator and ultimately getting the $h$ in the leading $\dfrac{1}{h}$ factor to divide out with an $h$ factor in the numerator.by the same token, let $\Delta t = h$ in the second limit; also, using the properties of limits will make the overall task a bit easier ...
$\displaystyle x’ = 3 \cdot \lim_{h \to 0} \dfrac{(t+h)^2-t^2}{h} - 2 \cdot \lim_{h \to 0} \dfrac{\sqrt{t+h} - \sqrt{t}}{h}$
the limit of the first term is straightforward; the second will require you to rationalize the numerator
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