How to Find The Derivative of A Function
Derivatives, a cornerstone of calculus, reveal how functions change at specific points. This article explores calculating derivatives using limits, a fundamental method demonstrating how function values change as two points get infinitesimally close. We apply limits and algebraic techniques like conjugation to simplify and find the derivative. This approach offers a powerful tool for understanding the rate of change in various mathematical contexts.
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Understanding Derivatives Through Limits
We know that by finding the rate of change of the functions value between two points that are really close to each other, we find the slope or the derivative of that function at that point. Let’s express this mathematically, using the concept of limits:
\(\lim_{h \to 0} \frac{f(x+h) – f(x)}{(x+h) – x}\)
(\(h \to 0\) means \(h\) gets really small)
By plugging different functions in the limit above and some simplifying, we end up with general formulas or rules, so we don’t have to repeat similar calculation next time, for a similar function. Instead we plug into the rules and find the derivatives that way. If you follow the derivative rules closely, you’ll find that the answers will be the same when finding the derivatives using rules, or limits.
Practical Examples of Derivatives
Lets consider a simple function to see how this limit works in action:
\(f(x)=x\) or: \(y=x\)
For this function, \(f(x+h)=x+h\), so:
\(\lim_{h \to 0} \frac{x+h-x}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1\)
Therefore, the derivative of \(x\) is always gonna be \(1\).
This is the derivative with respect to \(x\).
So the derivative of \(x\) is always \(1\). Remember this for the next time you encounter \(y=x\) and need to find its derivative.
Advanced Example: Derivative of a Radical Function
Now, let’s try a slightly more complex function: \(f(x) = \sqrt{x + 1}\)
To find its derivative, we use the limit:
\(\lim_{h \to 0} \frac{\sqrt{x+h+1} – \sqrt{x+1}}{h}\)
, which looks complex at first, but we can multiply both numerator and denominator by a suitable \(1\), called conjugate. This exploits the algebraic identity: \((a-b)(a+b)=a^2-b^2\), to dissolve some of the radicals. The conjugate will be
\(\sqrt{x+h+1} + \sqrt{x+1}\),
but divided by itself to basically be equal to \(1\). So:
\( \frac{{\sqrt{x+h+1}} – \sqrt{x+1}}{h} \cdot \frac{\sqrt{x+h+1} + \sqrt{x+1}}{\sqrt{x+h+1} + \sqrt{x+1}} \)
\(= \frac{\sqrt{x+h+1} – \sqrt{x+1}}{\sqrt{x+h+1} + \sqrt{x+1}} \cdot {h(\sqrt{x+h+1} + \sqrt{x+1})} \)
\(= \frac{(x+h+1) – (x+1)}{h(\sqrt{x+h+1} + \sqrt{x+1})} \)
\(= \frac{h}{h(\sqrt{x+h+1} + \sqrt{x+1})} \)
\(= \frac{1}{\sqrt{x+h+1} + \sqrt{x+1}} \)
Back to the limit:
\(\lim_{h \to0}\frac {1}{\sqrt(x+h+1)+\sqrt(x+1)}\)
\(=\frac{1}{\sqrt(x+1)+\sqrt(x+1)}=\frac{1}{2\sqrt(x+1)}=\frac{1}{(2f(x)}\)
So similar radical derivatives can be calculated using this formula. But not all radicals.
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