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I'm trying to differentiate $x\sqrt{4-x^2}$ using the definition of derivative.

So it would be something like

$$\underset{h\to 0}{\text{lim}}\frac{(h+x) \sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}$$

I was trying to solve and I just can end up with something like

$$\underset{h\to 0}{\text{lim}}\frac{(x+h)\sqrt{4-x^2-2xh-h^2}-x\sqrt{4-x^2}}h \cdot \frac{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$

$$\underset{h\to 0}{\text{lim}}\frac{-3x^2h-3xh^2+4h-h^3+\sqrt{4-x^2}-\sqrt{4-x^2-2xh+h^2}}{h\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$

Now if I group on h, I will have some tricky 3 instead of 2. The idea is I should have something like $h(2x^2+4)$ that would cancel up.

I'm quite stuck can I ask a little of help? I know wolframalpha exists but it refuses to create the step by step solution with the error "Ops we don't have a step by step solution for this query".

The final result shall be $$-\frac{2 \left(x^2-2\right)}{\sqrt{4-x^2}}$$

2'5 9'2
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Delayer
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  • Use L'Hospital's rule to solve the limit. You can easily get the final answer. – User 123 Feb 12 '22 at 18:12
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    If I can use L'hospital than I would directly use the differentiation rules, what if it's not allowed? Not solvable? @User123 The exercise says "use the definition" – Delayer Feb 12 '22 at 18:16

3 Answers3

3

I think your algebra could look more like:

$$\begin{align} &\frac{(x+h)\sqrt{4-(x+h)^2}-x\sqrt{4-x^2}}{h}\cdot\frac{(x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}}{(x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}}\\ &=\frac{(x+h)^2(4-(x+h)^2)-x^2(4-x^2)}{h\left((x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}\right)}\\ \end{align}$$

This leaves no radicals in the numerator.

In the numerator, once this is multiplied out, all $h$-free terms will have canceled out through adding terms to their negatives.

Then you can factor $h$ from the top and cancel the $h$ in the denominator. Then it will be OK to just let $h\to0$.

2'5 9'2
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1

As Spivak points out in his Calculus, the proofs of limit theorems are strategies for implementing the definition of a limit.

Here, guided by the trick \begin{align*} \frac{f(x + h)g(x + h) - f(x)g(x)}{h} &= \frac{f(x + h)g(x + h) - f(x + h)g(x) + f(x + h)g(x) - f(x)g(x)}{h} \\ &= f(x + h)\frac{g(x + h) - g(x)}{h} + \frac{f(x + h) - f(x)}{h}g(x) \end{align*} in the proof of the power rule, we can add and subtract from the numerator either of $$ f(x + h)g(x) = (x + h)\sqrt{4 - x^{2}},\qquad f(x)g(x + h) = x\sqrt{4 - (x^{2} + 2xh + h^{2})}. $$ (Alex's (+1) answer is another approach, but the one here applies more generally.)

Andrew D. Hwang
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Clearly, when $h\to 0$: $$\frac{h\cdot \sqrt{4-\left(h^2+2 h x+x^2\right)}}{h}\to \sqrt{4-x^2}$$ So, the limit simplify to: $$\lim_{h\to 0}\frac{x\sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}=\sqrt{4-x^2}+\lim_{h\to 0}\frac{x\left[\sqrt{4-\left(h^2+2 h x+x^2\right)}-\sqrt{4-x^2}\right]}{h}=\sqrt{4-x^2}+\lim_{h\to 0}\frac{x\cdot \sqrt{4-x^2}\left[\sqrt{1-\frac{h^2+2hx}{4-x^2}}-1\right]}{h}$$

Using the fact that: $$\sqrt{1+x}\,\, \sim\,\, \frac{1}{2}x\,\, x \to 0$$ We obtain: $$\lim_{h\to 0}\frac{x\cdot \sqrt{4-x^2}\left[\sqrt{1-\frac{h^2+2hx}{4-x^2}}-1\right]}{h}\,\, \sim\,\, \lim_{h\to 0}\frac{x\cdot \sqrt{4-x^2}\cdot\left(-\frac{1}{2}\cdot\frac{2hx}{4-x^2}\right)}{h}=\lim_{h\to 0}\frac{-\frac{hx^2}{\sqrt{4-x^2}}}{h}=-\frac{x^2}{\sqrt{4-x^2}}$$

So, the limit is: $$\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}=\frac{4-x^2-x^2}{\sqrt{4-x^2}}=2\cdot\frac{2-x^2}{\sqrt{4-x^2}}$$

Matteo
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