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प्रश्न
Differentiate the following w.r.t.x: log[tan3x.sin4x.(x2 + 7)7]
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उत्तर
Let y = log[tan3x.sin4x.(x2 + 7)7]
= log tan3x + log sin4x + log(x2 + 7)7
= 3log tanx + 4log sinx + 7log(x2 + 7)
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[3log tanx + 4 logsinx + 7 log(x^2 + 7)]`
= `3"d"/"dx"(log tan x) + 4"d"/"dx"(log sinx) + 7"d"/"dx"[log(x^2 + 7)]`
= `3 xx (1)/tanx ."d"/"dx"(tanx) + 4 xx (1)/sinx."d"/"dx"(sinx) + 7 xx (1)/(x^2 + 7)."d"/"dx"(x^2 + 7)`
= `3 xx (1)/tanx.sec^2x + 4 xx (1)/sinx.cosx + 7 xx (1)/(x^2 + 7).(2x + 0)`
= `3 xx "cosx"/"sinx" xx (1)/(cos^2x) + 4cotx + (14x)/(x^2 + 7)`
= `(6)/(2sinx cosx) + 4cot + (14x)/(x^2 + 7)`
= `(6)/(sin2x) + 4cotx + (14x)/(x^2 + 7)`
= `6"cosec"2x + 4cotx + (14x)/(x^2 + 7)`.
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