find g' for g=sqrt[tan(sqrt(x))+x) + sqrt(x^2+sin(e^(2x)))]

[jjedlicka]

Help with take home test: I need help finding the derivative of this fuction

$\displaystyle g(x)=sqrt{tan((sqrt({x}))+x)+sqrt{(x^2+sin(e^{2x}))}}$

 

[Deleted member 4993]

Re: find g' for g=sqrt[tan(sqrt(x))+x) + sqrt(x^2+sin(e^(2x)

Help with take home test: I need help finding the derivative of this fuction

$\displaystyle g(x)=sqrt{tan((sqrt({x}))+x)+sqrt{(x^2+sin(e^{2x}))}}$

This is a test - are you supposed to seek help??

Show your work and excatly where you are stuck - we might be able to "unstuck" you without compromising the test.

 

[jjedlicka]

Its a pratice test...
I get the answer
$\displaystyle (tan(x^{1/2}+x)+(x^2+sin(e^{2x}))^{1/2})^{1/2}$

$\displaystyle 1/2(tan(x^{1/2}+x)+(x^2+sin(e^{2x}))^{1/2})^{-1/2}$

$\displaystyle 1/2(sec^2(x^{1/2}+x)+1/2(x^2+sin(e^{2x})) ^{-1/2})^{-1/2}$

$\displaystyle 1/2(sec^2(x^{1/2}+x)+1/2(2x+cos(e^{2x}))^{-1/2})^{-1/2}$

$\displaystyle g'(x)=1/2(sec^2(1/2x^{-1/2}+1)+1/2(2x+cos(e^{2x})2xe^{2x})^{-1/2})^{-1/2}$

 

[Deleted member 4993]

Its a pratice test...
I get the answer
$\displaystyle (tan(x^{1/2}+x)+(x^2+sin(e^{2x}))^{1/2})^{1/2}$

$\displaystyle 1/2(tan(x^{1/2}+x)+(x^2+sin(e^{2x}))^{1/2})^{-1/2}$

$\displaystyle 1/2(sec^2(x^{1/2}+x)+1/2(x^2+sin(e^{2x})) ^{-1/2})^{-1/2}$

$\displaystyle 1/2(sec^2(x^{1/2}+x)+1/2(2x+cos(e^{2x}))^{-1/2})^{-1/2}$

$\displaystyle g'(x)=1/2(sec^2(1/2x^{-1/2}+1)+1/2(2x+cos(e^{2x})2xe^{2x})^{-1/2})^{-1/2}$

That is not correct - you need to apply cahin rule - several times.

 

[jjedlicka]

i figured it wasnt the right answer...how do I get the right one