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 82059 by jagoll last updated on 18/Feb/20

what is derivative of h = (√(ln(x))) by first principle method

$${what}\:{is}\:{derivative}\:{of}\:\:{h}\:=\:\sqrt{{ln}\left({x}\right)} \\ $$$${by}\:{first}\:{principle}\:{method}\: \\ $$

Answered by Henri Boucatchou last updated on 18/Feb/20

((dh(x))/dx)=(d(√(lnx))/dx) =(((dlnx)/dx)/(2(√(lnx)))) =(1/(2x(√(lnx))))

$$\frac{\mathrm{dh}\left(\mathrm{x}\right)}{\mathrm{dx}}=\frac{\mathrm{d}\sqrt{\mathrm{lnx}}}{\mathrm{dx}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{dlnx}}{\mathrm{dx}}}{\mathrm{2}\sqrt{\mathrm{lnx}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2x}\sqrt{\mathrm{lnx}}} \\ $$

Answered by mr W last updated on 18/Feb/20

h′=lim_(Δx→0) (((√(ln (x+Δx)))−(√(ln x)))/(Δx)) =lim_(Δx→0) (((√(ln x+ln (1+((Δx)/x))))−(√(ln x)))/(Δx)) =(√(ln x))lim_(Δx→0) (((√(1+((ln (1+((Δx)/x)))/(ln x))))−1)/(Δx)) =(√(ln x))lim_(Δx→0) (({1+(1/2)[((ln (1+((Δx)/x)))/(ln x))]+...}−1)/(Δx)) =(√(ln x))lim_(Δx→0) {(1/2)[((ln (1+((Δx)/x)))/(Δx ln x))]+...} =(√(ln x))lim_(Δx→0) {(1/2)[((ln (1+((Δx)/x))^(x/(Δx)) )/(xln x))]+...} =(√(ln x))×(1/2)×((ln e)/(xln x)) =(1/(2x(√(ln x))))

$${h}'=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{ln}\:\left({x}+\Delta{x}\right)}−\sqrt{\mathrm{ln}\:{x}}}{\Delta{x}} \\ $$$$=\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{ln}\:{x}+\mathrm{ln}\:\left(\mathrm{1}+\frac{\Delta{x}}{{x}}\right)}−\sqrt{\mathrm{ln}\:{x}}}{\Delta{x}} \\ $$$$=\sqrt{\mathrm{ln}\:{x}}\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\Delta{x}}{{x}}\right)}{\mathrm{ln}\:{x}}}−\mathrm{1}}{\Delta{x}} \\ $$$$=\sqrt{\mathrm{ln}\:{x}}\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\Delta{x}}{{x}}\right)}{\mathrm{ln}\:{x}}\right]+...\right\}−\mathrm{1}}{\Delta{x}} \\ $$$$=\sqrt{\mathrm{ln}\:{x}}\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\Delta{x}}{{x}}\right)}{\Delta{x}\:\mathrm{ln}\:{x}}\right]+...\right\} \\ $$$$=\sqrt{\mathrm{ln}\:{x}}\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\Delta{x}}{{x}}\right)^{\frac{{x}}{\Delta{x}}} }{{x}\mathrm{ln}\:{x}}\right]+...\right\} \\ $$$$=\sqrt{\mathrm{ln}\:{x}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{ln}\:{e}}{{x}\mathrm{ln}\:{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{\mathrm{ln}\:{x}}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com