$$\frac{\mathrm{d} }{\mathrm{d} x}x\sin \left ( \sqrt{3x^{2}+5} \right )$$
I can't for the life of me differentiate this function while only using Trig Identities, Basic differentiation rules, and Limits (no L'Hopital, either.)
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2$\begingroup$ Why do you want avoid chain rule? $\endgroup$ – user Oct 25 '20 at 7:41
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$\begingroup$ This is a challenge problem before my Unit 3 test of Derivatives in AP Calc AB. We haven't been taught the chain rule, and although I understand it, we can't use it. $\endgroup$ – Atticus Baker Oct 25 '20 at 7:56
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$$f'=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}$$ separate it into two limitation by $\pm(x)\sin\sqrt{3(x+t)^2+5}$ this means $$f'=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}=\\ =\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}\pm(x)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}\\ =\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x+t)^2+5}}{t}+\lim_{t \to 0}\frac{+x\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}$$
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$\begingroup$ Can I get the exact answer outside of Limit notation? $\endgroup$ – Atticus Baker Oct 25 '20 at 7:58
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Just add the proof of chain rule to your argument to have your cake and eat it too (but do not say His name). Of course, we can skip the $x$ in front by product rule for derivatives. Let $H(x) = \sqrt{3x^2+5}$. Then$^*$ \begin{align} \frac{\sin(H(x+t)) - \sin(H(x))}{t} &= \frac{\sin(H(x+t)) - \sin(H(x))}{t} \\ &= \frac{\sin(H(x+t)) - \sin(H(x))}{H(x+t)-H(x)}\frac{{H(x+t)-H(x)}}t \\ &= \frac{\sin( H(x)+[H(x+t)-H(x)] ) - \sin(H(x))}{H(x+t)-H(x)}\frac{{H(x+t)-H(x)}}t \end{align} Since $H(x+t)-H(x)\to 0$ as $t\to 0$, (which follows since $H$ is continuous), by the definition of a derivative and product rule for limits, we get \begin{align} \frac{d}{dx} \sin(H(x)) &= \lim_{t\to 0}\frac{\sin( H(x)+[H(x+t)-H(x)] ) - \sin(H(x))}{H(x+t)-H(x)}\frac{{H(x+t)-H(x)}}t \\ &=\lim_{t\to 0}\frac{\sin( H(x)+[H(x+t)-H(x)] ) - \sin(H(x))}{H(x+t)-H(x)}\lim_{t\to 0}\frac{{H(x+t)-H(x)}}t \\&= \cos(H(x)) H'(x). \end{align}
$^*$more words are needed if you want to worry (as you should) about dividing by $H(x+t)-H(x)$. But this is never zero, if we only look at $t$ sufficiently small, because $H$ is not locally constant.
answered Oct 25 '20 at 8:49
