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find the derivative f(x)=tan^(-1) (x/7)
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Do you know how to find the derivative of tan^(-1) ?
1/x^2+1
let's use the chain rule.
\[y = \tan^{-1}(u)\]
\[u = \frac{x}{7}\]
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we know that \[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\]
So \[\frac{dy}{dx}=\frac{1}{1+u^2}\frac{1}{7}\]
\[\frac{dy}{dx}=\frac{1}{1+(x/7)^2}\frac{1}{7}\]
Do you have any question?
its that whole thing right? 1/1+(x/7)^2*1/7???
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yes
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