Find $\frac{\mathrm{d} }{\mathrm{d} x}x\sin \left ( \sqrt{3x^{2}+5} \right )$ without using the chain rule. $$\frac{\mathrm{d} }{\mathrm{d} x}x\sin \left ( \sqrt{3x^{2}+5} \right )$$
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.)
 A: $$f'=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}$$ separate it into two limitation by $\pm(x)\sin\sqrt{3(x+t)^2+5}$
this means
$$f'=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}=\\
=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}\pm(x)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}\\
=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x+t)^2+5}}{t}+\lim_{t \to 0}\frac{+x\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}$$
A: 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.
