Question

If $f\left(x\right)=5^{g\left(x\right)}$ and $g\left(x\right)={\int }_2^{x^2}\frac{t}{\ln (1+t^2)}dt$, then find the value of $f^{\prime }\left(\sqrt{2}\right)$

Answer 1

Given

$g\left(x\right)={\int }_2^{x^2}\frac{t}{ln(1+t)}$

$g^{\prime }\left(x\right)=\frac{x^2}{ln(1+x^4)}\times 2x$

$g^{\prime }\left(x\right)=\frac{2x^3}{ln(1+x^4)}$

$f\left(x\right)=5^{g\left(x\right)}$

$\Rightarrow f^{\prime }\left(x\right)=5^{g\left(x\right)}.g^{\prime }\left(x\right).ln5$

$f^{\prime }\left(x\right)=5^{g\left(x\right)}.g^{\prime }\left(x\right).ln5$

$f^{\prime }\left(\sqrt{2}\right)=5^{g\left(\sqrt{2}\right)}.g^{\prime }\left(\sqrt{2}\right).ln5$

$=5^0\times \frac{2\times 2\sqrt{2}}{ln(1+4)}\times ln5$

$f^{\prime }\left(\sqrt{2}\right)=\frac{4\sqrt{2}}{ln5}\times ln5$

$\therefore f^{\prime }\left(\sqrt{2}\right)=4\sqrt{2}$