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Question
Find the derivative of f(x) = tan(ax + b), by first principle.
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Solution
We have f'(x) = `lim_(h -> 0) (f(x + h) - f(x))/h`
= `lim_(h -> 0) (tan(a(x + h) + b) - tan(ax + b))/h`
= `lim_(h -> 0) ((sin(ax + ah + b))/(cos(ax + ah + b)) - (sin(ax + b))/(cos(ax + b)))/h`
= `lim_(h -> 0) (sin(ax + ah + b) cos(ax + b) - sin(ax + b) cos(ax + ah + b))/(h cos(ax + b) cos(ax + ah + b))`
= `lim_(h -> 0) (a sin (ah))/(a * h cos (ax + b) cos(ax + ah + b))`
= `lim_(h -> 0) a/(cos(ax + b) cos(ax + ah + b))`
= `lim_(ah -> 0) (sin ah)/(ah)` ....[as h → 0 ah → 0]
= `a/(cos^2 (ax + b))`
= `a sec^2 (ax + b)`.
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