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प्रश्न
Differentiate the following w.r.t. x : `((x^2 + 2x + 2)^(3/2))/((sqrt(x) + 3)^3(cosx)^x`
उत्तर
Let y = `((x^2 + 2x + 2)^(3/2))/((sqrt(x) + 3)^3(cosx)^x`
Then log y = `log[((x^2 + 2x + 2)^(3/2))/((sqrt(x) + 3)^3(cosx)^x)]`
= `log(x^2 + 2x + 2)^(3/2) - log(sqrt(x) + 3)^3(cosx)^x`
= `(3)/(2)log(x^2 + 2x + 2) - 3log(sqrt(x) + 3) - xlog(cosx)`
Differentiating both sides w.r.t. x, we get
`(1)/y."dy"/"dx" = (3)/(2)"d"/"dx"[log(x^2 + 2x + 2)] -3"d"/"dx"[log(sqrt(x) + 3)] - "d"/"dx"[xlog(cosx)]`
= `(3)/(2) xx (1)/(x^2 + 2x + 2)."d"/"dx"(x^2 + 2x + 2) -3 xx (1)/(sqrt(x) + 3)."d"/"dx"(sqrt(x) + 3) - {x"d"/"dx"[log(cosx)] + log(cosx)."d"/"dx"(x)}`
= `(3)/(2(x^2 + 2x + 2)) xx (2x + 2 xx 1 + 0) - (3)/(sqrt(x) + 3) xx (1/(2sqrt(x)) + 0) - {x xx (1)/cosx."d"/"dx"(cosx) + log(cosx) xx 1}`
= `(3(2x + 2))/(2(x^2 + 2x + 2)) - (3)/(2sqrt(x)(sqrt(x) + 3)) - {x xx (1)/cosx.(-sinx) + log(cosx)}`
∴ `"dy"/"dx" = y[(3(x + 1))/(x^2 + 2x + 2) - (3)/(2sqrt(x)(sqrt(x) + 3)) + xtanx - log(cosx)]`
= `((x^2 + 2x + 2)^(3/2))/((sqrt(x) + 3)^3(cosx)^x)[(3(x + 1))/(x^2 + 2x + 2) - (3)/(2sqrt(x)(sqrt(x) + 3)) + xtanx - log(cosx)]`.
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