Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 153963 by ZiYangLee last updated on 12/Sep/21

Given that f and g are differentiable  functions such that f ′(x)=(1/( (√(1+(f(x))^2 )) )),  and g=f^( −1)  , find g′(x).

$$\mathrm{Given}\:\mathrm{that}\:{f}\:\mathrm{and}\:{g}\:\mathrm{are}\:\mathrm{differentiable} \\ $$$$\mathrm{functions}\:\mathrm{such}\:\mathrm{that}\:{f}\:'\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left({f}\left({x}\right)\right)^{\mathrm{2}} }\:}, \\ $$$$\mathrm{and}\:{g}={f}^{\:−\mathrm{1}} \:,\:\mathrm{find}\:{g}'\left({x}\right). \\ $$

Answered by mr W last updated on 12/Sep/21

y=f(x)  y′=(1/( (√(1+y^2 ))))  (√(1+y^2 ))dy=dx  (1/2)[ln (y+(√(1+y^2 )))+y(√(1+y^2 ))]+C=x  ⇒g(x)=f^(−1) (x)=(1/2)[ln (x+(√(1+x^2 )))+x(√(1+x^2 ))]+C  g′(x)=(1/2)[((1+(x/( (√(1+x^2 )))))/(x+(√(1+x^2 ))))+(√(1+x^2 ))+(x^2 /( (√(1+x^2 ))))]  =(1/2)[((x+(√(1+x^2 )))/(x+(√(1+x^2 ))))+1+2x^2 ](1/( (√(1+x^2 ))))  =(√(1+x^2 ))

$${y}={f}\left({x}\right) \\ $$$${y}'=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }} \\ $$$$\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }{dy}={dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)+{y}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right]+{C}={x} \\ $$$$\Rightarrow{g}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)+{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right]+{C} \\ $$$${g}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}+\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}+\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right]\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$

Commented by ZiYangLee last updated on 12/Sep/21

i found a better way..

$$\mathrm{i}\:\mathrm{found}\:\mathrm{a}\:\mathrm{better}\:\mathrm{way}.. \\ $$

Commented by mr W last updated on 12/Sep/21

yes, your way is best!

$${yes},\:{your}\:{way}\:{is}\:{best}! \\ $$

Answered by ZiYangLee last updated on 12/Sep/21

let g(x)=f^( −1) (x)=y                      f(y)=x                 f ′(x) (dy/dx) =1            (1/( (√(1+[f(y)]^2 )) )) (dy/dx) =1                             (dy/dx)= (√(1+[f(y)]^2 ))                       g ′(x)= (√(1+[f(y)]^2 ))                       g ′(x)= (√(1+x^2 )) _#

$${let}\:{g}\left({x}\right)={f}^{\:−\mathrm{1}} \left({x}\right)={y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({y}\right)={x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\:'\left({x}\right)\:\frac{{dy}}{{dx}}\:=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left[{f}\left({y}\right)\right]^{\mathrm{2}} }\:}\:\frac{{dy}}{{dx}}\:=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dy}}{{dx}}=\:\sqrt{\mathrm{1}+\left[{f}\left({y}\right)\right]^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{g}\:'\left({x}\right)=\:\sqrt{\mathrm{1}+\left[{f}\left({y}\right)\right]^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{g}\:'\left({x}\right)=\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:_{#} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com