Question

If $f^{\prime }\left(x\right)=\sqrt{3x^2+4}$ and $y=f\left(x^3\right)$, then at $x=2$, $\frac{dy}{dx}$ equals __________.

Answer 1

Given , $f^{{}^{\prime }}\left(x\right)=\sqrt{{3x}^2+4}\&y=f\left(x^3\right)$

$\Rightarrow \frac{dy}{dx}=f^{{}^{\prime }}\left(x^3\right).3x^2=\sqrt{{3\left(x^3\right)}^2+4}.3x^2$

at $x=2,\frac{dy}{dx}=\sqrt{3{\left(8\right)}^2+4}.3.4=12\sqrt{196}=12\times 14=168$