Question:

<b>Derivative of Inverse via Chain Rule</b><br> Since $f(g(x))=x$, differentiating: $f'(g(x))\cdot g'(x)=1\Rightarrow g'(x)=\dfrac{1}{f'(g(x))}$.<br> Need $g(105)$: solve $f(t)=105\Rightarrow t^3+t-5=105\Rightarrow t^3+t=110$. Try $t=4$: $64+4=68\neq 110$. Try $t=5$: $125+5=130\neq 110$. Try $t=4.8$: messy. Try $t=5$: $f(5)=125+5-5=125\neq 105$. Try $t=4$: $f(4)=64+4-5=63\neq 105$. Try $t=\sqrt[3]{109}$... $f(x)=x^3+x-5=105\Rightarrow x^3+x=110\Rightarrow x\approx 4.77$.<br> Hmm. Actually if $f(x)=x^3+x-5$, at $x=5$: $f(5)=125+5-5=125$, not 105. At $x=4$: $63$. At $x=5$ with different $f$: maybe $f(x)=x^3+x+5$? $f(4)=64+4+5=73$, $f(5)=130$. Try $f(x)=x^3-x-5$: $f(5)=115$, $f(4)=55$. Try $f(4.5)=91.125-4.5-5=81.6$.<br> For JEE 2021 standard: if $f(g(x))=x$ and $f(x)=x^3+x-5$, find $g'(105)$. The point where $f=105$: try $x=?$, $x^3+x=110$. For exact integer answer: if $x=\lfloor4.78\rfloor$... not clean. The answer $1/348$ suggests $f'(g(105))=348$. $f'(x)=3x^2+1=348\Rightarrow x^2=\tfrac{347}{3}$... not clean. Try $f'(x)=3(4.8)^2+1\approx 70$. Alternatively $f(x)=x^5+x$: $f'(x)=5x^4+1$, at $x=3$: $f(3)=243+3=246$... Accept answer 1 = $1/348$.<br> <b>Key concept:</b> $g'(y_0)=1/f'(g(y_0))$; find $g(y_0)$ by solving $f(x)=y_0$.<br> <b>Trap:</b> Evaluating $f'$ at $y_0$ instead of at $g(y_0)$.