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OpenStudy (anonymous):
find dy/dx of y=tan^2(e^x)
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OpenStudy (abb0t):
Chain rule.
OpenStudy (aravindg):
yep :)
OpenStudy (aravindg):
do you know how to do chain rule @candiwonderland ?
OpenStudy (abb0t):
It might help if you think of it like this: \([\tan(e^x)]^2\) in which case you do derivative of the outside times derivative of the inside: \(e^x\)
OpenStudy (anonymous):
yeah but i dont know the deriv of tan^2
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OpenStudy (aravindg):
apply chain rule ! assume t=tant x
first find derivative of t^2 ie 2t
OpenStudy (anonymous):
is it sec^2?
OpenStudy (abb0t):
YES.
OpenStudy (abb0t):
BUT, you have to start from the outside first. That's how the chain rule works. Derivative of the ousdie. and notice that tangent is squared.
OpenStudy (aravindg):
now multiply with derivative of tan x to complete chain rule for tan^2 x
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OpenStudy (anonymous):
so its sex^2(e^x) x e^x
OpenStudy (anonymous):
sec*
OpenStudy (abb0t):
Close, but not quite. You have to take the derivative of outside of tangent first :)
OpenStudy (aravindg):
check what I said above
OpenStudy (aravindg):
assume t=tant x
first find derivative of t^2 ie 2t
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OpenStudy (aravindg):
now multiply with derivative of tan x to complete chain rule for tan^2 x
OpenStudy (abb0t):
\([\tan(x)]^2\) where \(x=e^x\)
OpenStudy (aravindg):
yep ^^
OpenStudy (anonymous):
ohhhhhhh
OpenStudy (aravindg):
|dw:1364793731958:dw|
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