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Differentiate the following w.r.t.x: log (sec 3x+ tan 3x) - Mathematics and Statistics

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प्रश्न

Differentiate the following w.r.t.x:

log (sec 3x+ tan 3x)

योग
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उत्तर

Let y = log (sec 3x+ tan 3x)
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[log (sec 3x+ tan 3x)]`

= `(1)/(sec 3x + tan 3x)."d"/"dx"(sec 3x + tan 3x)`

= `(1)/(sec 3x + tan 3x) xx ["d"/"dx"(sec3x) + "d"/"dx"(tan 3x)]`

= `(1)/(sec 3x + tan 3x) xx [sec3x tan3x. "d"/"dx"(3x) + sec^2 3x."d"/"dx"(3x)]`

= `(1)/(sec 3x + tan 3x) xx [sec 3x tan3x xx 3 + sec^2 3x xx 3]`

= `(3sec 3x(tan3x + sec3x))/(sec 3x + tan3x)`
= 3sec 3x

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Differentiation - Exercise 1.1 [पृष्ठ १२]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.1 | Q 3.07 | पृष्ठ १२

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