Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 2035 by Rasheed Soomro last updated on 31/Oct/15

f( f ′(x) )=f ′( f(x) )  f(x)=?

$${f}\left(\:{f}\:'\left({x}\right)\:\right)={f}\:'\left(\:{f}\left({x}\right)\:\right) \\ $$$${f}\left({x}\right)=? \\ $$

Commented by Yozzi last updated on 31/Oct/15

Try f(x)=e^x . f^′ (x)=e^x ⇒f(f^′ (x))=e^e^x    f^′ (f(x))=e^e^x    f(x)=a^x ⇒f^′ (x)=a^x lna (a>0)  ∴f(f^′ (x))=a^(a^x lna)   f^′ (f(x))=a^(a^x lna) lna≠f(f^′ (x))

$${Try}\:{f}\left({x}\right)={e}^{{x}} .\:{f}^{'} \left({x}\right)={e}^{{x}} \Rightarrow{f}\left({f}^{'} \left({x}\right)\right)={e}^{{e}^{{x}} } \\ $$$${f}^{'} \left({f}\left({x}\right)\right)={e}^{{e}^{{x}} } \\ $$$${f}\left({x}\right)={a}^{{x}} \Rightarrow{f}^{'} \left({x}\right)={a}^{{x}} {lna}\:\left({a}>\mathrm{0}\right) \\ $$$$\therefore{f}\left({f}^{'} \left({x}\right)\right)={a}^{{a}^{{x}} {lna}} \\ $$$${f}^{'} \left({f}\left({x}\right)\right)={a}^{{a}^{{x}} {lna}} {lna}\neq{f}\left({f}^{'} \left({x}\right)\right) \\ $$$$ \\ $$

Commented by 123456 last updated on 31/Oct/15

f•f^(−1) =f^(−1) •f=I  (if f is invertible)  f[f′(x)]=f′[f(x)]  f′(x)=f^(−1) {f′[f(x)]}  f(x)=(f′)^(−1) {f[f′(x)]}

$${f}\bullet{f}^{−\mathrm{1}} ={f}^{−\mathrm{1}} \bullet{f}=\mathrm{I}\:\:\left(\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{invertible}\right) \\ $$$${f}\left[{f}'\left({x}\right)\right]={f}'\left[{f}\left({x}\right)\right] \\ $$$${f}'\left({x}\right)={f}^{−\mathrm{1}} \left\{{f}'\left[{f}\left({x}\right)\right]\right\} \\ $$$${f}\left({x}\right)=\left({f}'\right)^{−\mathrm{1}} \left\{{f}\left[{f}'\left({x}\right)\right]\right\} \\ $$

Answered by 123456 last updated on 31/Oct/15

f(x)=a^x   f′(x)=a^x ln a          (a>0)  f[f′(x)]=a^(a^x ln a)   f′[f(x)]=a^a^x  ln a  a^(a^x ln a) =a^a^x  ln a        a^(a^x (ln a−1)) =ln a              (a^a^x  ≠0)  a^x (ln a−1)=log_a ln a  (ln a>0⇒a>1)  a^x =((log_a ln a)/(ln a−1))                     (ln a≠1⇒a≠e)  x=log_a (((log_a ln a)/(ln a−1)))  −−−−−−−−−−  a=e⇒ln a=1⇒log_a  ln a=0  f(x)=e^x    is a solution

$${f}\left({x}\right)={a}^{{x}} \\ $$$${f}'\left({x}\right)={a}^{{x}} \mathrm{ln}\:{a}\:\:\:\:\:\:\:\:\:\:\left({a}>\mathrm{0}\right) \\ $$$${f}\left[{f}'\left({x}\right)\right]={a}^{{a}^{{x}} \mathrm{ln}\:{a}} \\ $$$${f}'\left[{f}\left({x}\right)\right]={a}^{{a}^{{x}} } \mathrm{ln}\:{a} \\ $$$${a}^{{a}^{{x}} \mathrm{ln}\:{a}} ={a}^{{a}^{{x}} } \mathrm{ln}\:{a}\:\:\:\:\:\: \\ $$$${a}^{{a}^{{x}} \left(\mathrm{ln}\:{a}−\mathrm{1}\right)} =\mathrm{ln}\:{a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}^{{a}^{{x}} } \neq\mathrm{0}\right) \\ $$$${a}^{{x}} \left(\mathrm{ln}\:{a}−\mathrm{1}\right)=\mathrm{log}_{{a}} \mathrm{ln}\:{a}\:\:\left(\mathrm{ln}\:{a}>\mathrm{0}\Rightarrow{a}>\mathrm{1}\right) \\ $$$${a}^{{x}} =\frac{\mathrm{log}_{{a}} \mathrm{ln}\:{a}}{\mathrm{ln}\:{a}−\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{ln}\:{a}\neq\mathrm{1}\Rightarrow{a}\neq{e}\right) \\ $$$${x}=\mathrm{log}_{{a}} \left(\frac{\mathrm{log}_{{a}} \mathrm{ln}\:{a}}{\mathrm{ln}\:{a}−\mathrm{1}}\right) \\ $$$$−−−−−−−−−− \\ $$$${a}={e}\Rightarrow\mathrm{ln}\:{a}=\mathrm{1}\Rightarrow\mathrm{log}_{{a}} \:\mathrm{ln}\:{a}=\mathrm{0} \\ $$$${f}\left({x}\right)={e}^{{x}} \:\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$

Answered by prakash jain last updated on 31/Oct/15

f(x)=x     ....(solution 1)  f ′(x)=1  f(f ′(x))=1  f ′(f(x))=1  f(x)=(x^2 /2)    ...(solution 2)  f ′(x)=x  f(f′(x))=f(x)=(x^2 /2)  f ′(f(x))=f ′((x^2 /2))=(x^2 /2)

$${f}\left({x}\right)={x}\:\:\:\:\:....\left({solution}\:\mathrm{1}\right) \\ $$$${f}\:'\left({x}\right)=\mathrm{1} \\ $$$${f}\left({f}\:'\left({x}\right)\right)=\mathrm{1} \\ $$$${f}\:'\left({f}\left({x}\right)\right)=\mathrm{1} \\ $$$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\:\:\:...\left({solution}\:\mathrm{2}\right) \\ $$$${f}\:'\left({x}\right)={x} \\ $$$${f}\left({f}'\left({x}\right)\right)={f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${f}\:'\left({f}\left({x}\right)\right)={f}\:'\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$

Commented by prakash jain last updated on 31/Oct/15

f(x)=(x^n /k)  f ′(x)=((nx^(n−1) )/k)  f(f′(x))=(((((nx^(n−1) )/k))^n )/k) = ((n^n x^(n(n−1)) )/k^(n+1) )  f ′(f(x))=((n((x^n /k))^(n−1) )/k) = ((nx^(n(n−1)) )/k^n )   ((nx^(n(n−1)) )/k^n )=((n^n x^(n(n−1)) )/k^(n+1) )  k=n^(n−1)   f(x)=(x^n /n^(n−1) )

$${f}\left({x}\right)=\frac{{x}^{{n}} }{{k}} \\ $$$${f}\:'\left({x}\right)=\frac{{nx}^{{n}−\mathrm{1}} }{{k}} \\ $$$${f}\left({f}'\left({x}\right)\right)=\frac{\left(\frac{{nx}^{{n}−\mathrm{1}} }{{k}}\right)^{{n}} }{{k}}\:=\:\frac{{n}^{{n}} {x}^{{n}\left({n}−\mathrm{1}\right)} }{{k}^{{n}+\mathrm{1}} } \\ $$$${f}\:'\left({f}\left({x}\right)\right)=\frac{{n}\left(\frac{{x}^{{n}} }{{k}}\right)^{{n}−\mathrm{1}} }{{k}}\:=\:\frac{{nx}^{{n}\left({n}−\mathrm{1}\right)} }{{k}^{{n}} } \\ $$$$\:\frac{{nx}^{{n}\left({n}−\mathrm{1}\right)} }{{k}^{{n}} }=\frac{{n}^{{n}} {x}^{{n}\left({n}−\mathrm{1}\right)} }{{k}^{{n}+\mathrm{1}} } \\ $$$${k}={n}^{{n}−\mathrm{1}} \\ $$$${f}\left({x}\right)=\frac{{x}^{{n}} }{{n}^{{n}−\mathrm{1}} }\:\:\: \\ $$

Commented by Rasheed Soomro last updated on 01/Nov/15

Excellent!  This is general solution  covers  f(x)=x,f(x)=(x^2 /2),f(x)=(x^3 /9)....(a whole caegary)

$$\mathcal{E}{xcellent}!\:\:{This}\:{is}\:{general}\:{solution}\:\:{covers} \\ $$$${f}\left({x}\right)={x},{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}},{f}\left({x}\right)=\frac{{x}^{\mathrm{3}} }{\mathrm{9}}....\left({a}\:{whole}\:{caegary}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com