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प्रश्न
Differentiate the following w.r.t.x:
log (sec 3x+ tan 3x)
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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