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Can you check my work please? derivative of 2tan(x/2)-5
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by letting u = (x/2) y=f(u)=2tan(u) dy/du=(2)(sec^2u) du/dx=(1) y=2sec^(u) y' = 2sec^2(x/2) ?
not quite
yeah... because the answer is only sec^2(x/2)... can you tell me where i went wrong?
you forgot to cancel out the 2
the 2? mmm, you mean the 2 and the x/2?
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wait a sec
ok
\[ u = \frac{x}{2} = \frac{du}{dx} = \frac{1}{2}\] \[y = 2 tan(u) - 5 => \frac{dy}{dx} = 2sec^{2}(x/2) * \frac{1}{2}\] \[ \frac{dy}{dx} = sec^{2}(x/2)\]
ok this is some chain rule which i cant remmeber
@Miracrown: what do you think?
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mim, i think a lot :P
ahhh yes yes. now i get it. :)
thanks :D gonna post another problem.
:)
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