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differentiate y=tan2 (x/3) help me...
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\[y = \tan^{2} (\frac{ x }{ 3 })\]
You have to use the chain rule twice.
First, let x/3=u
\[y=\tan^2u\]
Differentiate it like the following;\[\frac{ dy }{ dx }=\frac{ dy }{ du }*\frac{ du }{ dx }\]
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\[\frac{ dy }{ du }=2\tan u \sec^2u\]
\[\frac{ du }{ dx }=\frac{ 1 }{ 3 }\]
Therefore, \[\frac{ dy }{ dx }=\frac{ 2 }{ 3 }\tan u \sec^2 u=\frac{ 2 }{ 3 }\tan \frac{ x }{ 3 } \sec^2 \frac{ x }{ 3 }\]
Oh, and my bad. You only have to use the chain rule once, not twice.
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okay!! thank u very much.
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