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Find the derivative of f(x) = tan(ax + b), by first principle. - Mathematics

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प्रश्न

Find the derivative of f(x) = tan(ax + b), by first principle.

योग
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उत्तर

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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अध्याय 13: Limits and Derivatives - Solved Examples [पृष्ठ २३५]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 13 Limits and Derivatives
Solved Examples | Q 19 | पृष्ठ २३५

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