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find g' for g=sqrt[tan(sqrt(x))+x) + sqrt(x^2+sin(e^(2x)))]
- Thread starter jjedlicka
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D
Deleted member 4993
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Re: find g' for g=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 said: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.
Its a pratice test...
I get the answer
\(\displaystyle (tan(x^{1/2}+x)+(x^2+sin(e^{2x}))^{1/2})^{1/2}\)
\(\displaystyle 1/2(tan(x^{1/2}+x)+(x^2+sin(e^{2x}))^{1/2})^{-1/2}\)
\(\displaystyle 1/2(sec^2(x^{1/2}+x)+1/2(x^2+sin(e^{2x})) ^{-1/2})^{-1/2}\)
\(\displaystyle 1/2(sec^2(x^{1/2}+x)+1/2(2x+cos(e^{2x}))^{-1/2})^{-1/2}\)
\(\displaystyle g'(x)=1/2(sec^2(1/2x^{-1/2}+1)+1/2(2x+cos(e^{2x})2xe^{2x})^{-1/2})^{-1/2}\)
I get the answer
\(\displaystyle (tan(x^{1/2}+x)+(x^2+sin(e^{2x}))^{1/2})^{1/2}\)
\(\displaystyle 1/2(tan(x^{1/2}+x)+(x^2+sin(e^{2x}))^{1/2})^{-1/2}\)
\(\displaystyle 1/2(sec^2(x^{1/2}+x)+1/2(x^2+sin(e^{2x})) ^{-1/2})^{-1/2}\)
\(\displaystyle 1/2(sec^2(x^{1/2}+x)+1/2(2x+cos(e^{2x}))^{-1/2})^{-1/2}\)
\(\displaystyle g'(x)=1/2(sec^2(1/2x^{-1/2}+1)+1/2(2x+cos(e^{2x})2xe^{2x})^{-1/2})^{-1/2}\)
D
Deleted member 4993
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jjedlicka said: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.