Question Number 215809 by efronzo1 last updated on 18/Jan/25
![If f(x) = 2 + ∫_1 ^(−x^3 ) (√(2+u^2 )) du find the value of (d/dx) [f^(−1) (x)]_(x=2)](Q215809.png)
$$\:\:\:\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{2}\:+\:\underset{\mathrm{1}} {\overset{−\mathrm{x}^{\mathrm{3}} } {\int}}\sqrt{\mathrm{2}+\mathrm{u}^{\mathrm{2}} }\:\mathrm{du}\: \\ $$$$\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)\right]_{\mathrm{x}=\mathrm{2}} \\ $$
Answered by mr W last updated on 19/Jan/25
![f(x)=2+∫_1 ^(−x^3 ) (√(2+u^2 ))du ((df(x))/dx)=−3x^2 (√(2+x^6 )) f(−1)=2 ⇒f^(−1) (2)=−1 (d/dx)[f^(−1) (x)]_(x=2) =(1/((d/dx)[f(x)]_(x=−1) )) =−(1/(3(√3)))=−((√3)/9) ✓](Q215850.png)
$${f}\left({x}\right)=\mathrm{2}+\int_{\mathrm{1}} ^{−{x}^{\mathrm{3}} } \sqrt{\mathrm{2}+{u}^{\mathrm{2}} }{du} \\ $$$$\frac{{df}\left({x}\right)}{{dx}}=−\mathrm{3}{x}^{\mathrm{2}} \sqrt{\mathrm{2}+{x}^{\mathrm{6}} } \\ $$$${f}\left(−\mathrm{1}\right)=\mathrm{2}\:\Rightarrow{f}^{−\mathrm{1}} \left(\mathrm{2}\right)=−\mathrm{1} \\ $$$$\frac{{d}}{{dx}}\left[{f}^{−\mathrm{1}} \left({x}\right)\right]_{{x}=\mathrm{2}} =\frac{\mathrm{1}}{\frac{{d}}{{dx}}\left[{f}\left({x}\right)\right]_{{x}=−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\:\:\checkmark \\ $$