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Question Number 72343 by Rio Michael last updated on 27/Oct/19
given that y = ln ( 1 + cos^2 x) find (dy/(dx  )) at the point  x = ((3π)/4)  and  if  y =ln(x^2  + 4) find  (dy/dx) at x = 1
$${given}\:{that}\:{y}\:=\:{ln}\:\left(\:\mathrm{1}\:+\:{cos}^{\mathrm{2}} {x}\right)\:{find}\:\frac{{dy}}{{dx}\:\:}\:{at}\:{the}\:{point}\:\:{x}\:=\:\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$${and}\:\:{if}\:\:{y}\:={ln}\left({x}^{\mathrm{2}} \:+\:\mathrm{4}\right)\:{find}\:\:\frac{{dy}}{{dx}}\:{at}\:{x}\:=\:\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 27/Oct/19
y(x)=ln(1+cos^2 x) ⇒y^′ (x)=((−2sinx cosx)/(1+cos^2 x)) ⇒  y^′ (((3π)/4))=((−2sin(((3π)/4))cos(((3π)/4)))/(1+cos^2 (((3π)/4)))) =((2sin((π/4))cos((π/4)))/(1+cos^2 ((π/4))))=((2.(1/( (√2) )).(1/( (√2))))/(1+((1/( (√2))))^2 ))  =(1/(3/2))=(2/3)  y(x)=ln(x^2 +4) ⇒y^′ (x)=((2x)/(x^2  +4)) ⇒y^′ (1)=(2/5)
$${y}\left({x}\right)={ln}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)\:\Rightarrow{y}^{'} \left({x}\right)=\frac{−\mathrm{2}{sinx}\:{cosx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:\Rightarrow \\ $$$${y}^{'} \left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)=\frac{−\mathrm{2}{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right){cos}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}\:=\frac{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)}=\frac{\mathrm{2}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{\mathrm{1}+\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{2}}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${y}\left({x}\right)={ln}\left({x}^{\mathrm{2}} +\mathrm{4}\right)\:\Rightarrow{y}^{'} \left({x}\right)=\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow{y}^{'} \left(\mathrm{1}\right)=\frac{\mathrm{2}}{\mathrm{5}} \\ $$
Commented by Rio Michael last updated on 30/Oct/19
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 31/Oct/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$

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